Stoichiometry & Chemical Calculations - DAT Survey of the Natural Sciences

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)
Answer: 2.0 moles 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.
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)
Answer: 3.6 × 10²⁴ atoms 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.
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)
Answer: 50% 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.
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)
Answer: 50.0 g 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.
A 1.80 g sample of a compound containing only carbon, hydrogen, and oxygen is combusted completely, producing 2.64 g of CO₂ and 1.08 g of H₂O. What is the empirical formula of the compound?
Answer: CH₂O When you encounter combustion analysis problems, you're determining empirical formulas by working backwards from the products formed when a compound burns completely in oxygen. The key insight is that all carbon atoms end up in CO₂ and all hydrogen atoms end up in H₂O. Start by finding moles of each element. From 2.64 g CO₂: $\frac{2.64 \text{ g}}{44.01 \text{ g/mol}} = 0.0600 \text{ mol CO₂}$ , which means 0.0600 mol carbon atoms. From 1.08 g H₂O: $\frac{1.08 \text{ g}}{18.02 \text{ g/mol}} = 0.0599 \text{ mol H₂O}$ , which means 0.120 mol hydrogen atoms (since each H₂O contains 2 H atoms). For oxygen, subtract the mass of C and H from the total sample mass. Carbon mass: 0.0600 mol × 12.01 g/mol = 0.721 g. Hydrogen mass: 0.120 mol × 1.008 g/mol = 0.121 g. Oxygen mass: 1.80 - 0.721 - 0.121 = 0.958 g, which equals $\frac{0.958}{16.00} = 0.0599$ mol oxygen atoms. The molar ratio is C:H:O = 0.0600:0.120:0.0599 ≈ 1:2:1, giving empirical formula CH₂O. This matches choice B. Choice A (CHO) has a 1:1:1 ratio, which doesn't match our 1:2:1 calculation. Choice C (CH₃O) suggests a 1:3:1 ratio. Choice D (C₂H₄O) reduces to CH₂O when simplified, but represents a different molecular formula entirely. Remember: always convert product masses to moles of elements first, then find the simplest whole-number ratio. Double-check by ensuring the calculated mass matches the given sample mass.
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)
Answer: 40.0 g NaOH 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.
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)
Answer: 2.0 g 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.
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?
Answer: 160.0 g 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.
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)
Answer: 13 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.
A chemical reaction has a calculated theoretical yield of 85.0 grams. If an experiment is performed and 68.0 grams of product are actually isolated, what is the percent yield of the reaction?
Answer: 80.0% Percent yield questions test your understanding of the relationship between theoretical and actual experimental results in chemistry. When you see a problem asking for percent yield, you're comparing what should happen theoretically to what actually happened in the lab. Percent yield is calculated using the formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$ In this problem, you have an actual yield of 68.0 grams and a theoretical yield of 85.0 grams. Substituting these values: $\text{Percent Yield} = \frac{68.0 \text{ g}}{85.0 \text{ g}} \times 100\% = 0.80 \times 100\% = 80.0\%$ Looking at the wrong answers: Choice A (17.0%) represents the difference between theoretical and actual yields (85.0 - 68.0 = 17.0), but this isn't how percent yield is calculated. Choice C (85.0%) incorrectly uses the theoretical yield as the answer, missing the calculation entirely. Choice D (125%) would only be possible if the actual yield exceeded the theoretical yield, which would suggest an error in the theoretical calculation or experimental contamination. Remember that percent yield in real experiments is almost always less than 100% due to side reactions, incomplete reactions, or product loss during isolation. When you see percent yield problems on the DAT, immediately identify your actual and theoretical yields, then apply the formula carefully, watching your decimal placement when converting to percentage.
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.
Answer: 59.4 g 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.
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)
Answer: 10.2 g 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.
How many moles of mercury (Hg) are present in a 10.0 mL sample? (Density of Hg = 13.6 g/mL; Molar Mass of Hg ≈ 200.6 g/mol)
Answer: 0.678 mol This question tests your ability to convert between different units using density and molar mass - a fundamental skill in chemistry that connects the physical properties of substances to their molecular composition. To solve this, you need to use dimensional analysis in two steps: first convert volume to mass using density, then convert mass to moles using molar mass. Start with the volume: 10.0 mL of mercury. Using the density (13.6 g/mL), calculate the mass: $\text{Mass} = 10.0 \text{ mL} \times 13.6 \text{ g/mL} = 136 \text{ g}$ Next, convert this mass to moles using mercury's molar mass (200.6 g/mol): $\text{Moles} = \frac{136 \text{ g}}{200.6 \text{ g/mol}} = 0.678 \text{ mol}$ This confirms answer C is correct. Looking at the wrong answers: A (0.050 mol) would result from incorrectly using 1.0 mL instead of 10.0 mL, or making a calculation error that's off by roughly a factor of 13. B (0.068 mol) appears to come from using an incorrect molar mass or making an arithmetic mistake in the division step. D (1.47 mol) likely results from incorrectly multiplying instead of dividing by the molar mass, or using the reciprocal of the density. Remember the conversion pathway for these problems: Volume → Mass (using density) → Moles (using molar mass). Set up your dimensional analysis carefully, ensuring units cancel properly at each step. This systematic approach prevents the calculation errors that create most wrong answer choices.
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)
Answer: 0.300 mol 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.
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?
Answer: CH₂O 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.
The combustion of a 5.00 g sample of a hydrocarbon (a compound containing only C and H) produces 15.7 g of CO₂ and 6.43 g of H₂O. What is the empirical formula of the hydrocarbon?
Answer: CH₂ When you encounter combustion analysis problems, you're determining the empirical formula by finding the moles of each element from the combustion products. The key insight is that all carbon becomes CO₂ and all hydrogen becomes H₂O. First, find moles of carbon from CO₂: $\frac{15.7 \text{ g CO}_2}{44.01 \text{ g/mol}} = 0.357 \text{ mol CO}_2$ . Since each CO₂ contains one carbon atom, you have 0.357 mol of carbon. Next, find moles of hydrogen from H₂O: $\frac{6.43 \text{ g H}_2\text{O}}{18.02 \text{ g/mol}} = 0.357 \text{ mol H}_2\text{O}$ . Since each H₂O contains two hydrogen atoms, you have $0.357 \times 2 = 0.714 \text{ mol hydrogen}$ . To find the empirical formula, determine the simplest whole number ratio by dividing by the smallest number of moles: Carbon ratio = $\frac{0.357}{0.357} = 1$ ; Hydrogen ratio = $\frac{0.714}{0.357} = 2$ . This gives CH₂, which is answer choice A. Choice B (CH) would require equal moles of carbon and hydrogen, but you calculated twice as much hydrogen. Choice C (CH₃) would need a 1:3 carbon-to-hydrogen ratio, requiring 1.071 mol of hydrogen. Choice D (C₂H₅) represents a 2:5 ratio, which doesn't match your 1:2 ratio when reduced to simplest terms. Remember: always convert combustion products to moles of elements first, then find the simplest whole number ratio. Don't work directly with mass ratios.
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)
Answer: 111.7 g 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.
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)
Answer: 46.7% 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.
A 10.00 g sample of hydrated copper(II) sulfate, CuSO₄·xH₂O, is heated until all the water is driven off. The remaining anhydrous salt weighs 6.40 g. What is the integer value of x in the formula?
Answer: 5 When you encounter hydrated salt problems, you're dealing with stoichiometry and the concept that heating drives off water of crystallization. The key is to find how many moles of water were lost per mole of anhydrous salt. First, calculate the mass of water driven off: 10.00 g - 6.40 g = 3.60 g of water lost. Next, find the moles of each component. The anhydrous copper(II) sulfate (CuSO₄) has a molar mass of 159.6 g/mol, so: $\frac{6.40 \text{ g}}{159.6 \text{ g/mol}} = 0.0401 \text{ mol CuSO₄}$ The water lost has a molar mass of 18.0 g/mol: $\frac{3.60 \text{ g}}{18.0 \text{ g/mol}} = 0.200 \text{ mol H₂O}$ To find x, determine the mole ratio: $\frac{0.200 \text{ mol H₂O}}{0.0401 \text{ mol CuSO₄}} = 4.99 ≈ 5$ Therefore, x = 5, making answer C correct. Let's examine why the other options are wrong. Choice A (x = 2) would predict only 1.44 g of water, far less than the 3.60 g actually lost. Choice B (x = 3) would predict 2.16 g of water, still too low. Choice D (x = 7) would predict 5.04 g of water, which exceeds what was actually driven off. Remember for hydration problems: always work with mole ratios, not just mass ratios. Calculate the moles of anhydrous salt and water separately, then find their ratio to determine the number of water molecules in the original hydrated compound.
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)
Answer: 149 g/mol 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.

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