DAT Survey of the Natural Sciences Quiz: Stoichiometry And Chemical Calculations

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Stoichiometry And Chemical CalculationsQuestion 1 of 20

What is the molar mass of ammonium phosphate, (NH₄)₃PO₄? (Atomic masses: N≈14.0, H≈1.0, P≈31.0, O≈16.0 g/mol)

113 g/mol

121 g/mol

132 g/mol

149 g/mol

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DAT Survey of the Natural Sciences Quiz

DAT Survey of the Natural Sciences Quiz: Stoichiometry And Chemical Calculations

Practice Stoichiometry And Chemical Calculations in DAT Survey of the Natural Sciences with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Stoichiometry And Chemical Calculations, giving you a quick way to practice the rules, question types, and explanations that matter most for DAT Survey of the Natural Sciences.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the molar mass of ammonium phosphate, (NH₄)₃PO₄? (Atomic masses: N≈14.0, H≈1.0, P≈31.0, O≈16.0 g/mol)

113 g/mol
121 g/mol
132 g/mol
149 g/mol (correct answer)

Explanation: When calculating molar mass, you need to account for every atom in the molecular formula by multiplying each element's atomic mass by how many times it appears, then sum everything together. For ammonium phosphate

(NH_4)_3PO_4

, break down the formula systematically. The subscript 3 outside the parentheses means you have three ammonium ions, so you're dealing with: 3 nitrogen atoms, 12 hydrogen atoms (3 × 4), 1 phosphorus atom, and 4 oxygen atoms. Now calculate:

Nitrogen: $3 × 14.0 = 42.0$ g/mol
Hydrogen: $12 × 1.0 = 12.0$ g/mol
Phosphorus: $1 × 31.0 = 31.0$ g/mol
Oxygen: $4 × 16.0 = 64.0$ g/mol

Total:

42.0 + 12.0 + 31.0 + 64.0 = 149.0

g/mol Choice A (113 g/mol) likely comes from forgetting to multiply the ammonium group by 3, calculating as if there's only one

NH_4

. Choice B (121 g/mol) might result from miscounting hydrogen atoms or making arithmetic errors in the addition. Choice C (132 g/mol) could stem from incorrectly handling the polyatomic ions or using wrong atomic masses. The correct answer is D (149 g/mol). For molar mass problems, always write out exactly how many of each atom you have before calculating. Pay special attention to subscripts outside parentheses—they multiply everything inside. Double-check your arithmetic, as these problems test both your understanding of molecular formulas and your computational accuracy.

Question 2

When the following equation is balanced with the smallest whole-number coefficients, what is the sum of the coefficients for the reactants?

__ Cu + __ HNO₃ → __ Cu(NO₃)₂ + __ NO₂ + __ H₂O

3
5 (correct answer)
7
9

Explanation: When balancing chemical equations, you're applying the law of conservation of mass—atoms can't be created or destroyed, only rearranged. This redox reaction between copper and nitric acid requires systematic balancing. Start by identifying what's happening: copper is oxidized to Cu²⁺ while some nitrate is reduced to NO₂. Begin with the most complex molecule, Cu(NO₃)₂, which contains 2 nitrogen atoms. Since each HNO₃ provides one nitrogen, you'll need multiple HNO₃ molecules. Working through the balance systematically: Cu needs a coefficient of 1, but Cu(NO₃)₂ requires 2 nitrates for the copper salt plus additional HNO₃ to produce NO₂ and H₂O. The balanced equation is:

\text{Cu} + 4\text{HNO}_3 → \text{Cu(NO}_3\text{)}_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}

The reactant coefficients are 1 (for Cu) + 4 (for HNO₃) = 5. Choice A (3) likely comes from incorrectly counting only Cu + 2HNO₃, missing that you need 4 HNO₃ total. Choice C (7) probably includes products in the sum—remember the question asks only for reactant coefficients. Choice D (9) suggests adding all coefficients in the equation (1+4+1+2+2), but again, only reactants matter here. For DAT chemistry questions, always read carefully whether they want reactants, products, or total coefficients. Balance systematically by starting with the most complex molecule and checking that all elements balance before finalizing your answer.

Question 3

What mass of water (H₂O) is produced from the complete combustion of 8.0 grams of hydrogen gas (H₂)? The balanced equation is 2H₂(g) + O₂(g) → 2H₂O(l). (Molar masses: H₂ ≈ 2.0 g/mol, H₂O ≈ 18.0 g/mol)

18.0 g
36.0 g
72.0 g (correct answer)
144.0 g

Explanation: This is a stoichiometry problem that requires you to convert between masses of reactants and products using molar ratios from the balanced equation. When you see combustion problems asking for mass relationships, always start by identifying the mole-to-mole ratio from the balanced equation. From the equation 2H₂(g) + O₂(g) → 2H₂O(l), you can see that 2 moles of H₂ produce 2 moles of H₂O, giving a 1:1 molar ratio. Now convert your given mass to moles: 8.0 g H₂ ÷ 2.0 g/mol = 4.0 mol H₂. Since the ratio is 1:1, 4.0 mol H₂ produces 4.0 mol H₂O. Finally, convert moles of water to mass: 4.0 mol H₂O × 18.0 g/mol = 72.0 g H₂O. Looking at the wrong answers: Choice A (18.0 g) represents the mass of just 1 mole of water—you likely forgot to account for having 4 moles of H₂ as your starting point. Choice B (36.0 g) suggests you correctly found 4 moles of H₂ but mistakenly used a 2:1 ratio instead of 1:1, thinking 4 moles H₂ produces only 2 moles H₂O. Choice D (144.0 g) indicates you incorrectly applied a 1:2 ratio, assuming each mole of H₂ produces 2 moles of H₂O. Remember this three-step approach for stoichiometry: convert given mass to moles, apply the molar ratio from the balanced equation, then convert back to the desired mass units.

Question 4

How many moles are present in a sample containing 1.204 × 10²⁴ molecules of carbon dioxide (CO₂)? (Avogadro's number ≈ 6.022 × 10²³ mol⁻¹)

0.50 moles
2.0 moles (correct answer)
4.0 moles
22.0 moles

Explanation: When you encounter questions asking to convert between molecules and moles, you're working with one of chemistry's most fundamental relationships: Avogadro's number. This constant (6.022 × 10²³) represents how many particles are in exactly one mole of any substance. To solve this, you need the conversion formula: moles = number of molecules ÷ Avogadro's number. Substituting the given values:

\text{moles} = \frac{1.204 \times 10^{24}}{6.022 \times 10^{23}} = 2.0 \text{ moles}

This confirms answer B is correct. Let's examine why the other options are wrong. Answer A (0.50 moles) results from incorrectly multiplying instead of dividing, or from reversing the fraction setup. Answer C (4.0 moles) likely comes from calculation errors with scientific notation, perhaps doubling the correct answer due to arithmetic mistakes. Answer D (22.0 moles) suggests confusion with molar mass concepts or significant errors in handling the powers of 10. The key insight is recognizing that 1.204 × 10²⁴ is exactly twice Avogadro's number (since 1.204 is twice 0.602, and we're working with the same power of 10). This means you have exactly 2 moles. For DAT success, memorize Avogadro's number and practice scientific notation arithmetic. Always check if your answer makes intuitive sense—since the number of molecules given is about twice Avogadro's number, you should expect about 2 moles.

Question 5

A reaction is initiated with 5.4 g of aluminum (Al) and 6.4 g of oxygen (O₂). What mass of aluminum oxide (Al₂O₃) is formed? The balanced equation is 4Al + 3O₂ → 2Al₂O₃. (Molar masses: Al≈27, O₂≈32, Al₂O₃≈102 g/mol)

10.2 g (correct answer)
13.6 g
15.3 g
20.4 g

Explanation: This is a limiting reagent stoichiometry problem, where you need to determine which reactant runs out first and limits the amount of product formed. Start by converting grams to moles for each reactant. For aluminum:

\frac{5.4 \text{ g}}{27 \text{ g/mol}} = 0.2 \text{ mol Al}

. For oxygen:

\frac{6.4 \text{ g}}{32 \text{ g/mol}} = 0.2 \text{ mol O}_2

. Now use the balanced equation

4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3

to find the limiting reagent. From 0.2 mol Al, you could theoretically make

0.2 \text{ mol Al} \times \frac{2 \text{ mol Al}_2\text{O}_3}{4 \text{ mol Al}} = 0.1 \text{ mol Al}_2\text{O}_3

. From 0.2 mol O₂, you could make

0.2 \text{ mol O}_2 \times \frac{2 \text{ mol Al}_2\text{O}_3}{3 \text{ mol O}_2} = 0.133 \text{ mol Al}_2\text{O}_3

. Since aluminum produces less product, it's the limiting reagent. Convert 0.1 mol Al₂O₃ to grams:

0.1 \text{ mol} \times 102 \text{ g/mol} = 10.2 \text{ g}

, which is answer A. Answer B (13.6 g) comes from incorrectly using oxygen as the limiting reagent. Answer C (15.3 g) likely results from adding the masses of reactants and subtracting some arbitrary amount. Answer D (20.4 g) represents twice the correct answer, suggesting an error in the stoichiometric ratios. Remember: in limiting reagent problems, always calculate how much product each reactant could theoretically produce, then choose the smaller amount. The limiting reagent determines your maximum yield.

Question 6

A 10.0 g mixture of calcium carbonate (CaCO₃) and sodium chloride (NaCl) is treated with excess hydrochloric acid. Only the CaCO₃ reacts, producing 2.2 g of carbon dioxide (CO₂) gas. What was the mass percentage of CaCO₃ in the original mixture? (Molar masses: CO₂≈44, CaCO₃≈100 g/mol)

22%
44%
50% (correct answer)
78%

Explanation: This is a stoichiometry problem involving a mixture analysis where only one component reacts. When you see questions about mixtures with selective reactions, work backwards from the product to find the reactive component. Start with the balanced chemical equation:

\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2

. This shows that 1 mole of CaCO₃ produces 1 mole of CO₂. Since 2.2 g of CO₂ was produced, calculate the moles:

\frac{2.2 \text{ g}}{44 \text{ g/mol}} = 0.05 \text{ mol CO}_2

. Because of the 1:1 stoichiometric ratio, 0.05 mol of CaCO₃ must have been present in the mixture. Convert this to mass:

0.05 \text{ mol} \times 100 \text{ g/mol} = 5.0 \text{ g CaCO}_3

The mass percentage is:

\frac{5.0 \text{ g}}{10.0 \text{ g}} \times 100\% = 50\%

Answer A (22%) represents the percentage that CO₂ mass is of the total mixture mass, which confuses product with reactant. Answer B (44%) is the molar mass of CO₂, showing confusion between molar mass and percentage. Answer D (78%) might result from calculation errors or misunderstanding the stoichiometry. The correct answer is C (50%). Study tip: In mixture problems with selective reactions, always work backwards from the measured product to find the amount of the reactive component. The unreactive component (NaCl here) is just a distractor—focus on the stoichiometry between the reactive compound and its product.

Question 7

What mass of sodium hydroxide (NaOH) is required to completely neutralize 49.0 g of sulfuric acid (H₂SO₄)? The balanced equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. (Molar masses: H₂SO₄ ≈ 98.0 g/mol, NaOH ≈ 40.0 g/mol)

20.0 g NaOH
40.0 g NaOH (correct answer)
49.0 g NaOH
80.0 g NaOH

Explanation: When you encounter stoichiometry problems involving acid-base neutralization, you need to use the balanced chemical equation to determine the molar relationships between reactants, then convert between grams and moles using molar masses. Start by finding the moles of sulfuric acid:

\frac{49.0 \text{ g H}_2\text{SO}_4}{98.0 \text{ g/mol}} = 0.500 \text{ mol H}_2\text{SO}_4

The balanced equation shows that 1 mole of H₂SO₄ requires 2 moles of NaOH for complete neutralization. This 1:2 ratio is crucial because sulfuric acid is diprotic (releases two H⁺ ions), while sodium hydroxide is monoprotic (accepts one H⁺ ion). Using stoichiometry:

0.500 \text{ mol H}_2\text{SO}_4 \times \frac{2 \text{ mol NaOH}}{1 \text{ mol H}_2\text{SO}_4} = 1.00 \text{ mol NaOH}

Convert moles of NaOH to grams:

1.00 \text{ mol NaOH} \times 40.0 \text{ g/mol} = 40.0 \text{ g NaOH}

This confirms answer choice B is correct. Choice A (20.0 g) represents using a 1:1 molar ratio, ignoring that H₂SO₄ is diprotic. Choice C (49.0 g) assumes equal masses are needed, which ignores the different molar masses and stoichiometry. Choice D (80.0 g) doubles the correct answer, perhaps from incorrectly applying the 2:1 ratio in the wrong direction. Remember: always identify the mole ratio from the balanced equation first, then use dimensional analysis to convert grams → moles → moles → grams. Pay special attention to polyprotic acids requiring multiple equivalents of base.

Question 8

How many moles of oxygen gas (O₂) are produced from the thermal decomposition of 24.5 g of potassium chlorate (KClO₃)? The balanced equation is 2KClO₃(s) → 2KCl(s) + 3O₂(g). (Molar mass KClO₃ ≈ 122.5 g/mol)

0.100 mol
0.200 mol
0.300 mol (correct answer)
0.400 mol

Explanation: This is a stoichiometry problem involving mass-to-mole conversions and mole ratios from a balanced chemical equation. When you see decomposition reactions with given masses, always work through three steps: convert mass to moles of the given substance, use the balanced equation to find the mole ratio, then calculate moles of the desired product. First, convert the given mass of KClO₃ to moles:

\frac{24.5 \text{ g KClO₃}}{122.5 \text{ g/mol}} = 0.200 \text{ mol KClO₃}

Next, use the balanced equation's mole ratio. The equation shows that 2 moles of KClO₃ produce 3 moles of O₂, giving us a ratio of

\frac{3 \text{ mol O₂}}{2 \text{ mol KClO₃}}

Finally, calculate:

0.200 \text{ mol KClO₃} \times \frac{3 \text{ mol O₂}}{2 \text{ mol KClO₃}} = 0.300 \text{ mol O₂}

Looking at the wrong answers: Choice A (0.100 mol) likely comes from incorrectly using a 1:1 ratio and dividing by 2. Choice B (0.200 mol) assumes a 1:1 mole ratio between KClO₃ and O₂, ignoring the coefficients in the balanced equation. Choice D (0.400 mol) might result from multiplying by 2 instead of the correct factor of 1.5. Remember: stoichiometry problems always require you to use the coefficients from the balanced equation as conversion factors. Never assume 1:1 ratios unless the coefficients are actually equal. Practice identifying the given substance, the desired substance, and the mole ratio between them.

Question 9

A mixture of 10.0 g of hydrogen gas (H₂) and 64.0 g of oxygen gas (O₂) is ignited. What mass of the excess reactant remains after the reaction is complete? The balanced equation is 2H₂ + O₂ → 2H₂O. (Molar masses: H₂≈2.0, O₂≈32.0 g/mol)

2.0 g (correct answer)
8.0 g
16.0 g
32.0 g

Explanation: This is a limiting reagent problem that tests your ability to determine which reactant runs out first and calculate how much excess reactant remains unused. Start by converting grams to moles for both reactants. You have 10.0 g H₂ ÷ 2.0 g/mol = 5.0 mol H₂, and 64.0 g O₂ ÷ 32.0 g/mol = 2.0 mol O₂. Next, use the balanced equation

2H_2 + O_2 → 2H_2O

to determine the limiting reagent. The stoichiometry shows you need 2 moles of H₂ for every 1 mole of O₂. With 2.0 mol O₂ available, you would need 2.0 mol O₂ × 2 = 4.0 mol H₂. Since you have 5.0 mol H₂ available but only need 4.0 mol, oxygen is the limiting reagent and hydrogen is in excess. Calculate the excess hydrogen: 5.0 mol H₂ available - 4.0 mol H₂ consumed = 1.0 mol H₂ remaining. Converting back to grams: 1.0 mol × 2.0 g/mol = 2.0 g H₂. Looking at the wrong answers: B) 8.0 g might result from incorrectly calculating 10.0 g - 2.0 g instead of finding the actual excess. C) 16.0 g and D) 32.0 g likely come from misidentifying the limiting reagent or making calculation errors with the oxygen masses. The correct answer is A) 2.0 g. Study tip: Always identify the limiting reagent first by comparing mole ratios from the balanced equation, then calculate what's left over from the excess reactant.

Question 10

How many grams of oxygen gas (O₂) are required for the complete combustion of 50.0 mL of liquid methanol (CH₃OH)? (Density of CH₃OH = 0.792 g/mL; Molar masses: CH₃OH ≈ 32.0 g/mol, O₂ ≈ 32.0 g/mol) The balanced equation is 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O.

29.7 g
39.6 g
59.4 g (correct answer)
79.2 g

Explanation: This is a stoichiometry problem that requires you to convert between volume, mass, and moles using the balanced chemical equation. When you see combustion problems with given volumes and densities, always work systematically: volume → mass → moles → stoichiometry → final answer. Start by converting the volume of methanol to mass:

50.0 \text{ mL} \times 0.792 \text{ g/mL} = 39.6 \text{ g CH₃OH}

Next, convert mass to moles:

39.6 \text{ g} \div 32.0 \text{ g/mol} = 1.24 \text{ mol CH₃OH}

Now use the balanced equation's mole ratio. The equation shows that 2 moles of CH₃OH react with 3 moles of O₂, so the ratio is 3:2. Calculate moles of O₂ needed:

1.24 \text{ mol CH₃OH} \times \frac{3 \text{ mol O₂}}{2 \text{ mol CH₃OH}} = 1.86 \text{ mol O₂}

Finally, convert to grams:

1.86 \text{ mol} \times 32.0 \text{ g/mol} = 59.4 \text{ g O₂}

Answer choice A (29.7 g) represents using the wrong stoichiometric ratio—likely using 1:1 instead of 3:2. Answer choice B (39.6 g) is a trap that gives you the mass of methanol instead of oxygen. Answer choice D (79.2 g) appears to double the correct answer, possibly from miscalculating the mole ratio as 3:1. Remember: stoichiometry problems always follow the same pattern—convert to moles, apply the balanced equation ratios, then convert back to the desired units. Double-check your mole ratios from the balanced equation.

Question 11

The decomposition of 130 g of sodium azide (NaN₃) is used to inflate airbags via the reaction 2NaN₃(s) → 2Na(s) + 3N₂(g). Approximately how many individual nitrogen atoms (N) are produced? (Molar mass NaN₃ ≈ 65 g/mol; Avogadro's number ≈ 6.0 × 10²³ mol⁻¹)

1.8 × 10²⁴ atoms
2.4 × 10²⁴ atoms
3.6 × 10²⁴ atoms (correct answer)
7.2 × 10²⁴ atoms

Explanation: This stoichiometry problem tests your ability to convert between mass, moles, molecules, and atoms using dimensional analysis. When you see questions asking for "individual atoms" from a chemical reaction, you'll need to track the stoichiometry carefully and use Avogadro's number. Start by finding moles of sodium azide:

\frac{130 \text{ g NaN}_3}{65 \text{ g/mol}} = 2.0 \text{ mol NaN}_3

From the balanced equation

2\text{NaN}_3(s) \rightarrow 2\text{Na}(s) + 3\text{N}_2(g)

, you can see that 2 moles of NaN₃ produce 3 moles of N₂ gas. So 2.0 mol NaN₃ produces:

2.0 \text{ mol NaN}_3 \times \frac{3 \text{ mol N}_2}{2 \text{ mol NaN}_3} = 3.0 \text{ mol N}_2

Here's the crucial step: each N₂ molecule contains 2 nitrogen atoms. Therefore:

3.0 \text{ mol N}_2 \times 2 \text{ atoms N per molecule} = 6.0 \text{ mol N atoms}

Converting to individual atoms:

6.0 \text{ mol N} \times 6.0 \times 10^{23} \text{ atoms/mol} = 3.6 \times 10^{24} \text{ atoms}

Answer A (1.8 × 10²⁴) represents calculating only half the nitrogen atoms—forgetting that N₂ is diatomic. Answer B (2.4 × 10²⁴) likely comes from incorrectly using the original 2.0 mol without accounting for stoichiometry. Answer D (7.2 × 10²⁴) suggests doubling the correct answer, possibly from confusion about the balanced equation coefficients. For stoichiometry problems, always write out your dimensional analysis step-by-step and pay special attention to diatomic molecules when counting individual atoms.

Question 12

The complete decomposition of a sample of solid calcium carbonate (CaCO₃) produces 0.50 moles of carbon dioxide gas according to the reaction: CaCO₃(s) → CaO(s) + CO₂(g). What was the initial mass of the calcium carbonate sample? (Molar mass CaCO₃ ≈ 100.0 g/mol)

22.0 g
28.0 g
50.0 g (correct answer)
100.0 g

Explanation: This question tests stoichiometry - the quantitative relationship between reactants and products in chemical reactions. When you see a decomposition reaction with given product amounts, you need to work backward to find the initial reactant mass. The balanced equation shows a 1:1 molar ratio between CaCO₃ and CO₂. Since 0.50 moles of CO₂ were produced, exactly 0.50 moles of CaCO₃ must have decomposed. To find the mass, you multiply moles by molar mass:

0.50 \text{ mol} \times 100.0 \text{ g/mol} = 50.0 \text{ g}

Looking at the wrong answers: Choice A (22.0 g) likely comes from incorrectly using CO₂'s molar mass (≈44 g/mol) instead of CaCO₃'s molar mass. Choice B (28.0 g) doesn't correspond to any logical calculation error with the given values. Choice D (100.0 g) represents a common mistake where students confuse the molar mass value (100.0 g/mol) with the actual mass needed, forgetting to account for the 0.50 moles. The correct answer is C (50.0 g) because stoichiometry requires you to use the molar ratio from the balanced equation, then convert moles to grams using the correct compound's molar mass. For DAT stoichiometry problems, always follow this sequence: identify the molar relationship from the balanced equation, determine moles of the unknown substance, then convert using the appropriate molar mass. Double-check that you're using the molar mass of the compound you're solving for, not the given product.

Question 13

In the reduction of iron ore, 159.7 g of iron(III) oxide (Fe₂O₃) reacts with excess carbon monoxide (CO) according to the equation: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g). What is the theoretical yield of iron (Fe)? (Molar masses: Fe₂O₃ ≈ 159.7 g/mol, Fe ≈ 55.85 g/mol)

55.85 g
79.85 g
111.7 g (correct answer)
159.7 g

Explanation: When you encounter stoichiometry problems involving theoretical yield, you need to convert from the given substance to the desired product using molar ratios from the balanced equation. Start by identifying what you have and what you want. You have 159.7 g of Fe₂O₃ and need to find the theoretical yield of Fe. The balanced equation shows that 1 mole of Fe₂O₃ produces 2 moles of Fe. First, convert the mass of Fe₂O₃ to moles:

\frac{159.7 \text{ g Fe₂O₃}}{159.7 \text{ g/mol}} = 1.00 \text{ mol Fe₂O₃}

Next, use the molar ratio from the balanced equation. Since 1 mol Fe₂O₃ produces 2 mol Fe:

1.00 \text{ mol Fe₂O₃} × \frac{2 \text{ mol Fe}}{1 \text{ mol Fe₂O₃}} = 2.00 \text{ mol Fe}

Finally, convert moles of Fe to grams:

2.00 \text{ mol Fe} × 55.85 \text{ g/mol} = 111.7 \text{ g Fe}

Choice A (55.85 g) represents the mass of only 1 mole of Fe, ignoring that each Fe₂O₃ produces 2 Fe atoms. Choice B (79.85 g) appears to be an arithmetic error, possibly from incorrect molar mass calculations. Choice D (159.7 g) assumes a 1:1 mass ratio between reactant and product, which violates conservation of mass since other products form. The correct answer is C (111.7 g). For stoichiometry success, always follow the three-step pattern: convert given mass to moles, apply molar ratios from the balanced equation, then convert back to desired units.

Question 14

When the equation for the complete combustion of butane, C₄H₁₀, is balanced with the smallest whole-number coefficients, what is the coefficient for oxygen (O₂)?

__ C₄H₁₀(g) + __ O₂(g) → __ CO₂(g) + __ H₂O(g)

8
10
13 (correct answer)
25

Explanation: When you encounter combustion reactions, you're balancing a hydrocarbon with oxygen to produce carbon dioxide and water. The key is systematically balancing each element while ensuring you use the smallest whole-number coefficients. Start with the most complex molecule, butane (C₄H₁₀). Since it contains 4 carbon atoms, you need 4 CO₂ molecules on the product side. Since it contains 10 hydrogen atoms, you need 5 H₂O molecules (each water has 2 hydrogens). Your equation now looks like: C₄H₁₀ + __ O₂ → 4 CO₂ + 5 H₂O Now count the oxygen atoms needed on the product side: 4 CO₂ provides 8 oxygen atoms, and 5 H₂O provides 5 oxygen atoms, totaling 13 oxygen atoms. Since each O₂ molecule contains 2 oxygen atoms, you need 13/2 = 6.5 O₂ molecules. To get whole numbers, multiply the entire equation by 2: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O The coefficient for O₂ is 13. Looking at the wrong answers: A) 8 likely comes from only counting oxygen needed for CO₂ formation while forgetting about H₂O. B) 10 might result from confusing the coefficient with the number of hydrogen atoms in butane. D) 25 appears to be a distractor with no clear connection to the balanced equation. Study tip: Always balance combustion reactions in this order: carbon first, hydrogen second, then oxygen last. Remember that if you get fractional coefficients, multiply the entire equation by the appropriate factor to achieve whole numbers.

Question 15

In a reaction, 4.6 g of sodium (Na) is combined with 8.0 g of chlorine gas (Cl₂). Which is the limiting reactant? The balanced equation is 2Na(s) + Cl₂(g) → 2NaCl(s). (Molar masses: Na ≈ 23.0 g/mol, Cl₂ ≈ 71.0 g/mol)

Both reactants are completely consumed
Chlorine (Cl₂)
Sodium chloride (NaCl)
Sodium (Na) (correct answer)

Explanation: When you encounter stoichiometry problems asking about limiting reactants, you need to determine which reactant will be completely consumed first, stopping the reaction from proceeding further. To find the limiting reactant, convert the given masses to moles and compare them using the balanced equation's mole ratio. For sodium:

\frac{4.6 \text{ g}}{23.0 \text{ g/mol}} = 0.20 \text{ mol Na}

. For chlorine gas:

\frac{8.0 \text{ g}}{71.0 \text{ g/mol}} = 0.11 \text{ mol Cl}_2

. From the balanced equation 2Na + Cl₂ → 2NaCl, you can see that 2 moles of Na react with 1 mole of Cl₂. This means Na and Cl₂ react in a 2:1 ratio. With 0.11 mol of Cl₂ available, you would need

0.11 \times 2 = 0.22 \text{ mol Na}

. Since you only have 0.20 mol Na available, sodium will be consumed first, making it the limiting reactant. Choice A is incorrect because sodium runs out before chlorine is completely consumed. Choice B is wrong because chlorine is actually in excess—you have more than enough to react with all the available sodium. Choice C represents a product, not a reactant, so it cannot be a limiting reactant by definition. The correct answer is D—sodium is the limiting reactant. Remember: always convert to moles first, then use the balanced equation's coefficients to determine the required mole ratios. The reactant that runs out first limits how much product can form.

Question 16

A compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. What is the empirical formula of this compound?

CHO
C₂H₆O
C₂H₄O₂
CH₂O (correct answer)

Explanation: When you encounter a percent composition problem, you're being asked to find the simplest whole-number ratio of atoms in a compound. The key is converting mass percentages to moles, then finding the smallest whole-number ratio. Start by assuming you have 100g of the compound, so the percentages become grams: 40.0g C, 6.7g H, and 53.3g O. Convert each to moles using atomic masses (C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol):

Carbon: $\frac{40.0g}{12.01 g/mol} = 3.33 mol$
Hydrogen: $\frac{6.7g}{1.008 g/mol} = 6.65 mol$
Oxygen: $\frac{53.3g}{16.00 g/mol} = 3.33 mol$

Divide each by the smallest number of moles (3.33) to get the ratio:

C: $\frac{3.33}{3.33} = 1$
H: $\frac{6.65}{3.33} = 2$
O: $\frac{3.33}{3.33} = 1$

This gives the empirical formula CH₂O, which is answer D. Answer A (CHO) would result from incorrectly assuming a 1:1:1 ratio without doing the calculation. Answer B (C₂H₆O) represents a molecular formula that doesn't match our mass percentages. Answer C (C₂H₄O₂) is a multiple of the correct empirical formula but doesn't represent the simplest ratio. Remember: empirical formulas always show the simplest whole-number ratio. When your mole ratios aren't whole numbers, divide by the smallest value to reduce to the simplest form.

Question 17

What is the percent by mass of nitrogen in urea, CO(NH₂)₂? (Atomic masses: C≈12, O≈16, N≈14, H≈1 g/mol)

23.3%
31.8%
40.0%
46.7% (correct answer)

Explanation: When you encounter percent composition problems, you're finding what fraction of a compound's total mass comes from a specific element. This requires calculating the molecular mass of the entire compound, then determining how much of that mass comes from the element in question. First, find urea's molecular mass by adding all atomic masses. Urea, CO(NH₂)₂, contains: 1 carbon (12 g/mol), 1 oxygen (16 g/mol), 2 nitrogens (2 × 14 = 28 g/mol), and 4 hydrogens (4 × 1 = 4 g/mol). The total molecular mass is 12 + 16 + 28 + 4 = 60 g/mol. Next, calculate nitrogen's contribution. Since urea has 2 nitrogen atoms, nitrogen contributes 28 g/mol to the total mass. The percent by mass of nitrogen is:

\frac{28 \text{ g/mol}}{60 \text{ g/mol}} \times 100\% = 46.7\%

This confirms answer choice D is correct. Choice A (23.3%) likely results from counting only one nitrogen atom instead of two, giving you 14/60 × 100%. Choice B (31.8%) might come from calculation errors in the molecular mass or mixing up which atoms to count. Choice C (40.0%) could result from miscalculating the total molecular mass as 70 instead of 60, then using 28/70. For percent composition success, always double-check your molecular formula to count atoms correctly, especially when you see subscripts outside parentheses like the "2" in (NH₂)₂. This multiplies everything inside the parentheses.

Question 18

What is the approximate mass in grams of a single atom of silicon (Si)? (Molar mass of Si = 28.09 g/mol; Avogadro's number ≈ 6.022 × 10²³ mol⁻¹)

1.66 × 10⁻²⁴ g
4.66 × 10⁻²³ g (correct answer)
9.31 × 10⁻²³ g
28.09 g

Explanation: This question tests your understanding of the relationship between molar mass, Avogadro's number, and atomic mass at the individual atom level. When you see a problem asking for the mass of a single atom, you need to convert from the molar scale (which deals with enormous quantities) down to the individual particle level. To find the mass of one silicon atom, divide the molar mass by Avogadro's number:

\frac{28.09 \text{ g/mol}}{6.022 × 10^{23} \text{ atoms/mol}} = 4.66 × 10^{-23} \text{ g}

. This calculation converts from grams per mole to grams per atom, giving you answer B. Looking at the wrong choices: Answer A (1.66 × 10⁻²⁴ g) is approximately six times too small - this suggests an error in the calculation, possibly dividing by an incorrect version of Avogadro's number. Answer C (9.31 × 10⁻²³ g) is roughly double the correct value, indicating a computational mistake or unit error. Answer D (28.09 g) is simply the molar mass itself, representing the mass of one entire mole of silicon atoms rather than a single atom - this is a classic trap for students who forget to apply Avogadro's number. Remember this pattern: to find the mass of a single atom, always divide molar mass by Avogadro's number. The result will always be an extremely small number (around 10⁻²² to 10⁻²⁴ grams for most elements), so if your answer isn't in that range, double-check your calculation.

Question 19

When 10.0 g of a metal carbonate, MCO₃, is heated, 5.6 g of the metal oxide, MO, and 4.4 g of CO₂ are produced. What is the approximate molar mass of the metal M? (Atomic masses: C≈12, O≈16 g/mol)

24 g/mol
40 g/mol (correct answer)
56 g/mol
64 g/mol

Explanation: When you encounter stoichiometry problems involving decomposition reactions, you need to use the balanced chemical equation and molar relationships to find unknown quantities. This question tests your ability to work backwards from product masses to determine an unknown atomic mass. The decomposition reaction is:

\text{MCO}_3 \rightarrow \text{MO} + \text{CO}_2

. First, calculate moles of CO₂ produced:

\frac{4.4 \text{ g}}{44 \text{ g/mol}} = 0.10 \text{ mol}

. Since the reaction has a 1:1:1 molar ratio, 0.10 mol of MCO₃ decomposed to produce 0.10 mol of MO. Now you can find the molar mass of MCO₃:

\frac{10.0 \text{ g}}{0.10 \text{ mol}} = 100 \text{ g/mol}

. Since MCO₃ has a molar mass of 100 g/mol, and the carbonate portion (CO₃) contributes 60 g/mol (12 + 48), the metal M must have:

100 - 60 = 40 \text{ g/mol}

. You can verify this using MO:

\frac{5.6 \text{ g}}{0.10 \text{ mol}} = 56 \text{ g/mol}

for MO, so

56 - 16 = 40 \text{ g/mol}

for M. Choice A (24 g/mol) would give MCO₃ a mass of 84 g/mol, too low for the given data. Choice C (56 g/mol) represents the molar mass of MO, not M alone. Choice D (64 g/mol) would make MCO₃ have a mass of 124 g/mol, inconsistent with our calculations. For decomposition problems, always start by finding moles of a product with known molecular weight (like CO₂), then use stoichiometry to work backwards to the unknown compound.

Question 20

The synthesis of sulfur trioxide (SO₃) from sulfur dioxide (SO₂) and oxygen (O₂) is known to have a 75.0% yield. If a chemist obtains 120.0 g of SO₃, what was the theoretical yield for the reaction?

90.0 g
120.0 g
160.0 g (correct answer)
200.0 g

Explanation: When you encounter percent yield problems, you're working with the relationship between what actually happens in a reaction versus what theoretically should happen. The key formula is: % yield = (actual yield ÷ theoretical yield) × 100%. To find the theoretical yield, you need to rearrange this formula: theoretical yield = actual yield ÷ (% yield ÷ 100). Here, the actual yield is 120.0 g of SO₃ (what the chemist actually obtained), and the percent yield is 75.0%. Setting up the calculation: theoretical yield = 120.0 g ÷ (75.0% ÷ 100) = 120.0 g ÷ 0.750 = 160.0 g. This means if the reaction had proceeded with 100% efficiency, 160.0 g of SO₃ would have been produced, making C correct. Looking at the wrong answers: A (90.0 g) represents a common error where students multiply instead of divide: 120.0 g × 0.750 = 90.0 g. This gives you less than the actual yield, which is impossible since theoretical yield must be greater than actual yield when percent yield is less than 100%. B (120.0 g) incorrectly assumes the actual yield equals the theoretical yield, ignoring the 75% efficiency. D (200.0 g) might result from using the wrong percentage in calculations. Remember this pattern: when percent yield is less than 100%, the theoretical yield must always be larger than the actual yield. If your calculated theoretical yield is smaller than the given actual yield, you've made a calculation error.