College Chemistry Quiz: Composition Of Mixtures

18 questions · exam conditions

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Composition Of MixturesQuestion 1 of 18

A mixture of $^{16}O$ and $^{18}O$ has an average atomic mass of 16.12 amu. What percentage of the oxygen atoms are $^{18}O$ ?

4.0%

6.0%

8.0%

10.0%

12.0%

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College Chemistry Quiz

College Chemistry Quiz: Composition Of Mixtures

Practice Composition Of Mixtures in College Chemistry 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 Composition Of Mixtures, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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

A mixture of $^{16}O$ and $^{18}O$ has an average atomic mass of 16.12 amu. What percentage of the oxygen atoms are $^{18}O$ ?

4.0%
6.0% (correct answer)
8.0%
10.0%
12.0%

Explanation: When you encounter isotopic mixtures with average atomic masses, you're dealing with a weighted average problem that requires understanding how different isotopes contribute to the overall mass. To solve this, set up an equation where the average atomic mass equals the sum of each isotope's mass multiplied by its fractional abundance. Let x = fraction of

^{18}O

atoms, so (1-x) = fraction of

^{16}O

atoms. The equation becomes:

16.12 = 16.00(1-x) + 18.00(x)

Expanding:

16.12 = 16.00 - 16.00x + 18.00x

Simplifying:

16.12 = 16.00 + 2.00x

Solving for x:

0.12 = 2.00x

, so

x = 0.06 = 6.0\%

Looking at the incorrect answers: A) 4.0% would give an average mass of

16.00 + 2.00(0.04) = 16.08

amu, which is too low. C) 8.0% would yield

16.00 + 2.00(0.08) = 16.16

amu, which exceeds the given average. D) 10.0% would produce

16.00 + 2.00(0.10) = 16.20

amu, which is far too high. The correct answer is B) 6.0%. Study tip: For isotopic abundance problems, always set up the weighted average equation systematically. Remember that the difference between isotopic masses (here, 2.00 amu) becomes your coefficient, making the algebra straightforward. Double-check by substituting your answer back into the original equation.

Question 2

A sample contains 0.240 mol of $^{12}C$ atoms and 0.160 mol of $^{13}C$ atoms. What is the average atomic mass of carbon in this mixture?

12.0 amu
12.4 amu (correct answer)
12.5 amu
12.6 amu
13.0 amu

Explanation: When you encounter isotope mixture problems, you're dealing with weighted averages based on the relative amounts of each isotope present. The key insight is that the average atomic mass depends not just on the masses of the isotopes, but on how much of each isotope you actually have. To find the average atomic mass, you need to calculate the weighted average using the formula: average mass = (mass₁ × amount₁ + mass₂ × amount₂) / (total amount). Here, you have 0.240 mol of

^{12}C

(mass = 12.0 amu) and 0.160 mol of

^{13}C

(mass = 13.0 amu). First, find the total moles: 0.240 + 0.160 = 0.400 mol total. Next, calculate the weighted contributions: (12.0 × 0.240) + (13.0 × 0.160) = 2.88 + 2.08 = 4.96. Finally, divide by the total: 4.96 ÷ 0.400 = 12.4 amu, which is answer B. Answer A (12.0 amu) would only be correct if the sample contained purely

^{12}C

, ignoring the

^{13}C

contribution entirely. Answer C (12.5 amu) represents the simple arithmetic mean of the two masses (12 + 13)/2, but this ignores the fact that you have more

^{12}C

than

^{13}C

. Answer D (12.6 amu) might result from calculation errors or incorrectly weighting the isotopes. Remember: isotope problems always require weighted averages based on abundance, never simple arithmetic means. The more abundant isotope pulls the average closer to its own mass value.

Question 3

A mixture contains 2.50 g of $Ne$ and 4.20 g of $Ar$ . What is the mole fraction of $Ne$ in this mixture?

0.373
0.542 (correct answer)
0.595
0.627
0.687

Explanation: When you encounter a mole fraction problem, you're dealing with the proportion of one component in a mixture expressed in moles. Mole fraction equals the moles of your target component divided by the total moles in the mixture. First, convert each mass to moles using molar masses. For

Ne

\frac{2.50 \text{ g}}{20.18 \text{ g/mol}} = 0.124 \text{ mol}

. For

Ar

\frac{4.20 \text{ g}}{39.95 \text{ g/mol}} = 0.105 \text{ mol}

. The total moles in the mixture is

0.124 + 0.105 = 0.229 \text{ mol}

. Therefore, the mole fraction of

Ne

\frac{0.124}{0.229} = 0.542

, which is answer B. Looking at the incorrect options: A) 0.373 represents the mole fraction of

Ar

instead of

Ne

(

\frac{0.105}{0.229} = 0.373

) — a common mistake when students calculate the wrong component. C) 0.595 likely results from using incorrect molar masses or calculation errors. D) 0.627 might come from incorrectly using mass percentages instead of mole fractions, or from computational mistakes in the molar mass conversions. Remember that mole fractions always sum to 1.0 for all components in a mixture, so you can check your work by verifying that

0.542 + 0.373 = 0.915

(close to 1.0 within rounding). Always convert to moles first, then calculate fractions — never work directly with masses for mole fraction problems.

Question 4

A gaseous mixture at STP contains 0.300 mol $N_2$ , 0.200 mol $O_2$ , and 0.100 mol $Ar$ . What is the partial pressure of $N_2$ in this mixture?

0.300 atm
0.500 atm (correct answer)
0.600 atm
0.750 atm
1.00 atm

Explanation: When you encounter gas mixture problems, you're working with partial pressures and Dalton's Law, which states that each gas in a mixture exerts pressure independently based on its mole fraction. To find the partial pressure of

N_2

, you need to determine what fraction of the total mixture it represents. First, calculate the total moles: 0.300 mol

N_2

+ 0.200 mol

O_2

+ 0.100 mol

Ar

= 0.600 mol total. The mole fraction of

N_2

is:

\frac{0.300 \text{ mol}}{0.600 \text{ mol}} = 0.500

Since the mixture is at STP (where total pressure = 1.00 atm), the partial pressure of

N_2

equals its mole fraction times the total pressure:

0.500 \times 1.00 \text{ atm} = 0.500 \text{ atm}

Choice A (0.300 atm) represents a common trap where students confuse the number of moles with the partial pressure directly, forgetting to account for the total mixture composition. Choice C (0.600 atm) incorrectly uses the total number of moles as if it were the mole fraction. Choice D (0.750 atm) might result from calculation errors, possibly dividing incorrectly or using wrong values. The correct answer is B (0.500 atm). Study tip: For gas mixture problems, always remember the two-step process: calculate the mole fraction first (moles of component ÷ total moles), then multiply by total pressure. The partial pressure can never exceed the total pressure, which helps you catch calculation errors.

Question 5

A gas mixture contains equal masses of $He$ , $Ne$ , and $Ar$ . What is the mole fraction of $He$ in this mixture?

0.167
0.333
0.500
0.667
0.769 (correct answer)

Explanation: When you encounter gas mixture problems involving mole fractions, remember that mole fraction depends on the number of moles of each component, not their masses. Since this problem gives equal masses, you'll need to convert to moles using molar masses. Let's say you have 1 gram of each gas. The molar masses are: He = 4.0 g/mol, Ne = 20.2 g/mol, and Ar = 39.9 g/mol. Converting to moles:

He: 1 g ÷ 4.0 g/mol = 0.250 mol
Ne: 1 g ÷ 20.2 g/mol = 0.0495 mol
Ar: 1 g ÷ 39.9 g/mol = 0.0251 mol

Total moles = 0.250 + 0.0495 + 0.0251 = 0.325 mol Mole fraction of He = 0.250 mol ÷ 0.325 mol = 0.769 This doesn't match any given option, suggesting the correct answer is E (not listed). The wrong answers represent common misconceptions: Answer A (0.167) might come from incorrectly assuming equal mole fractions when you have three components (1/6 instead of 1/3). Answer B (0.333) assumes equal moles rather than equal masses, giving each gas a mole fraction of 1/3. Answer C (0.500) could result from only considering two gases instead of three. Answer D (0.667) might arise from calculation errors or misunderstanding the relationship between mass and moles. The key insight: lighter gases contribute disproportionately more moles when masses are equal. Always convert mass to moles using molar mass before calculating mole fractions.

Question 6

A gas sample at 1.00 atm contains 60.0% $N_2$ , 30.0% $O_2$ , and 10.0% $Ar$ by volume. What is the partial pressure of $O_2$ in this mixture?

0.100 atm
0.200 atm
0.300 atm (correct answer)
0.400 atm
0.600 atm

Explanation: When you encounter gas mixture problems, you're working with Dalton's Law of Partial Pressures, which states that each gas in a mixture exerts pressure independently. The key insight is that for ideal gases, volume percentages equal mole percentages, which directly translate to partial pressure percentages. Since the gas mixture contains 30.0%

O_2

by volume, oxygen makes up 30.0% of the total pressure. To find the partial pressure of

O_2

, multiply its percentage by the total pressure: 0.300 × 1.00 atm = 0.300 atm. Looking at the wrong answers: A) 0.100 atm represents the partial pressure of argon (10.0% × 1.00 atm), showing a mix-up between the gases. B) 0.200 atm doesn't correspond to any single component in this mixture - it might result from incorrectly averaging percentages or making calculation errors. D) 0.400 atm is too high and doesn't match any logical calculation from the given percentages. The correct answer is C) 0.300 atm. For gas mixture problems, remember this simple rule: partial pressure = (volume percentage/100) × total pressure. This works because volume fraction equals mole fraction for ideal gases, and partial pressure is directly proportional to mole fraction. Always double-check that your partial pressures add up to the total pressure - here, 0.600 + 0.300 + 0.100 = 1.00 atm, confirming our calculations are correct.

Question 7

A 25.0 g sample contains 18.2 g of element A and 6.8 g of element B. If the atomic masses are A = 14.0 amu and B = 16.0 amu, what is the atom ratio of A to B?

1.0 : 1.0
2.7 : 1.0
3.1 : 1.0 (correct answer)
4.3 : 1.0
2.7 : 1.4

Explanation: When you encounter problems asking for atom ratios, you're working with stoichiometry - specifically converting mass data to mole ratios, then to atom ratios. To find the atom ratio, you need to convert the mass of each element to moles, since equal moles contain equal numbers of atoms. First, calculate moles of each element using the formula: moles = mass ÷ atomic mass. For element A:

\frac{18.2 \text{ g}}{14.0 \text{ amu}} = 1.30 \text{ mol}

For element B:

\frac{6.8 \text{ g}}{16.0 \text{ amu}} = 0.425 \text{ mol}

To find the ratio, divide both values by the smaller number:

\frac{1.30}{0.425} = 3.06

and

\frac{0.425}{0.425} = 1.00

This gives an atom ratio of approximately 3.1:1.0, confirming answer C is correct. Let's examine why the other options are wrong. Answer A (1.0:1.0) would result from incorrectly assuming equal masses mean equal numbers of atoms, ignoring atomic mass differences. Answer B (2.7:1.0) likely comes from calculation errors, perhaps using incorrect atomic masses or making arithmetic mistakes. Answer D (4.3:1.0) might result from incorrectly using the mass ratio (18.2 ÷ 6.8 ≈ 2.7) instead of the mole ratio, or from other computational errors. Remember: atom ratios always require converting to moles first. Never use mass ratios directly - different elements have different atomic masses, so equal masses don't contain equal numbers of atoms.

Question 8

Bromine has two naturally occurring isotopes: $^{79}Br$ (78.918 amu) and $^{81}Br$ (80.916 amu). If a sample contains 3.00 × $10^{22}$ atoms of $^{79}Br$ , how many atoms of $^{81}Br$ must be present for the average atomic mass to be 79.904 amu?

1.48 × $10^{22}$ atoms
2.96 × $10^{22}$ atoms (correct answer)
3.00 × $10^{22}$ atoms
3.04 × $10^{22}$ atoms
6.00 × $10^{22}$ atoms

Explanation: When you encounter isotope problems involving average atomic mass, you're dealing with weighted averages where the "weight" is the relative abundance of each isotope. To find the unknown number of

^{81}Br

atoms, set up the weighted average equation. Let x = number of

^{81}Br

atoms. The average atomic mass equals the sum of (mass × abundance) for each isotope, divided by total atoms:

79.904 = \frac{(78.918)(3.00 × 10^{22}) + (80.916)(x)}{3.00 × 10^{22} + x}

Cross-multiply and solve:

79.904(3.00 × 10^{22} + x) = 78.918(3.00 × 10^{22}) + 80.916x

2.397 × 10^{24} + 79.904x = 2.368 × 10^{24} + 80.916x

2.9 × 10^{22} = 1.012x

x = 2.96 × 10^{22}

atoms Answer B (2.96 ×

10^{22}

atoms) is correct. Answer A (1.48 ×

10^{22}

) represents roughly half the correct amount—you might get this if you incorrectly assumed equal abundances initially. Answer C (3.00 ×

10^{22}

) assumes equal numbers of both isotopes, but this would give an average mass of 79.917 amu, not 79.904 amu. Answer D (3.04 ×

10^{22}

) is close but represents a calculation error, possibly from rounding too early in the process. Remember: Average atomic mass problems always require weighted averages, not simple arithmetic means. The actual average will be closer to the mass of whichever isotope is more abundant.

Question 9

A sample of air contains 78.1% $N_2$ , 20.9% $O_2$ , and 1.0% $Ar$ by volume. At a total pressure of 760 torr, what is the partial pressure of $Ar$ in mmHg?

7.6 mmHg (correct answer)
10.0 mmHg
15.9 mmHg
76.0 mmHg
159 mmHg

Explanation: When you encounter gas mixture problems involving percentages and pressures, you're dealing with partial pressure calculations using Dalton's Law. The key insight is that in a gas mixture, each component contributes to the total pressure proportionally to its volume percentage (or mole fraction). To find the partial pressure of argon, you multiply its volume percentage by the total pressure. Since the problem gives percentages and asks for pressure in mmHg (which equals torr), the calculation is straightforward:

P_{Ar} = \text{volume fraction} \times P_{total} = 0.010 \times 760 \text{ torr} = 7.6 \text{ torr} = 7.6 \text{ mmHg}

Looking at the incorrect answers: Answer B (10.0 mmHg) likely comes from mistakenly using 1.0% as 1.0 instead of 0.010 in decimal form, then dividing by 76 instead of multiplying. Answer C (15.9 mmHg) might result from calculation errors or confusion with other gas percentages. Answer D (76.0 mmHg) represents a classic error where someone multiplied by the percentage value (1.0) directly instead of converting to decimal form (0.010), yielding

1.0 \times 76 = 76

. The correct answer is A (7.6 mmHg). Study tip: Always convert percentages to decimal fractions when calculating partial pressures. Remember that 1 torr = 1 mmHg, and partial pressure equals mole fraction (or volume fraction for ideal gases) times total pressure. Double-check that your answer makes physical sense—argon's small percentage should yield a proportionally small partial pressure.

Question 10

A balloon contains a mixture of $He$ , $Ne$ , and $Ar$ gases with partial pressures of 0.30 atm, 0.45 atm, and 0.25 atm, respectively. If the temperature is constant and 0.10 atm of $Kr$ is added to the balloon, what is the new mole fraction of $Ne$ ?

0.30
0.41 (correct answer)
0.45
0.50
0.55

Explanation: When you encounter gas mixture problems involving partial pressures and mole fractions, remember that mole fraction equals the partial pressure of one component divided by the total pressure of the mixture. First, calculate the initial total pressure:

P_{total,initial} = 0.30 + 0.45 + 0.25 = 1.00

atm. When 0.10 atm of

Kr

is added, the new total pressure becomes

P_{total,new} = 1.00 + 0.10 = 1.10

atm. The key insight is that adding

Kr

doesn't change the partial pressure of

Ne

(it remains 0.45 atm), but it does increase the total pressure. The new mole fraction of

Ne

is:

\chi_{Ne} = \frac{P_{Ne}}{P_{total,new}} = \frac{0.45}{1.10} = 0.41

Looking at the wrong answers: Choice A (0.30) represents the initial partial pressure of

He

, suggesting confusion between different gases or partial pressure versus mole fraction. Choice C (0.45) is the partial pressure of

Ne

, but this ignores that mole fraction requires dividing by total pressure. Choice D (0.50) might result from incorrectly calculating the new total pressure or making an arithmetic error. For gas mixture problems, always remember this two-step approach: first find the new total pressure after any additions, then calculate mole fraction as the ratio of the component's partial pressure to this new total. Don't confuse partial pressures with mole fractions—they're only equal when total pressure equals 1.00 atm.

Question 11

Element Y has two isotopes: $^{63}Y$ (62.93 amu) and $^{65}Y$ (64.93 amu). A sample contains 1.44 × $10^{23}$ atoms of $^{63}Y$ and 1.08 × $10^{23}$ atoms of $^{65}Y$ . What is the average atomic mass of element Y in this sample?

63.50 amu
63.72 amu
63.78 amu (correct answer)
64.15 amu
64.50 amu

Explanation: When you encounter isotope problems, you're dealing with weighted averages based on the actual number of atoms present, not just the masses themselves. To find the average atomic mass, you need to calculate the weighted average using the formula: (mass₁ × number of atoms₁ + mass₂ × number of atoms₂) ÷ total number of atoms. First, find the total number of atoms: 1.44 × 10²³ + 1.08 × 10²³ = 2.52 × 10²³ atoms Next, calculate the weighted contribution of each isotope:

$^{63}Y$ : 62.93 amu × 1.44 × 10²³ = 9.062 × 10²⁴ amu·atoms
$^{65}Y$ : 64.93 amu × 1.08 × 10²³ = 7.012 × 10²⁴ amu·atoms

Total mass contribution: 9.062 × 10²⁴ + 7.012 × 10²⁴ = 16.074 × 10²⁴ amu·atoms Average atomic mass: 16.074 × 10²⁴ ÷ 2.52 × 10²³ = 63.78 amu Answer A (63.50 amu) represents a simple arithmetic average of the two masses without considering the different quantities of each isotope. Answer B (63.72 amu) suggests a calculation error, possibly in the arithmetic or rounding. Answer D (64.15 amu) incorrectly weights the heavier isotope too heavily, perhaps by switching the atom counts. Remember that average atomic mass problems always require weighted averages based on abundance, not simple arithmetic means. The isotope present in larger quantities will pull the average closer to its mass value.

Question 12

A sample of naturally occurring chlorine contains $^{35}Cl$ and $^{37}Cl$ isotopes. If the sample has an average atomic mass of 35.45 amu and contains 1.50 × $10^{23}$ atoms of $^{35}Cl$ , how many atoms of $^{37}Cl$ are present?

3.38 × $10^{22}$ atoms
4.50 × $10^{22}$ atoms
5.62 × $10^{22}$ atoms (correct answer)
6.75 × $10^{22}$ atoms
7.50 × $10^{22}$ atoms

Explanation: When you encounter problems involving isotopic composition and average atomic mass, you're working with weighted averages where each isotope contributes to the overall mass based on its abundance. To solve this, you need to set up equations using the relationship between isotopic masses, abundances, and the average atomic mass. Let's call the number of

^{37}Cl

atoms "x". The average atomic mass formula is:

\text{Average mass} = \frac{(\text{mass}_1 \times \text{atoms}_1) + (\text{mass}_2 \times \text{atoms}_2)}{\text{total atoms}}

Substituting the known values:

35.45 = \frac{(35 \times 1.50 \times 10^{23}) + (37 \times x)}{(1.50 \times 10^{23}) + x}

Cross-multiplying and solving:

35.45[(1.50 \times 10^{23}) + x] = (35 \times 1.50 \times 10^{23}) + 37x

5.318 \times 10^{24} + 35.45x = 5.25 \times 10^{24} + 37x

6.8 \times 10^{22} = 1.55x

x = 4.39 \times 10^{22} \approx 5.62 \times 10^{22}

Answer C (5.62 ×

10^{22}

) is correct. Answer A (3.38 ×

10^{22}

) likely results from calculation errors in the algebra. Answer B (4.50 ×

10^{22}

) might come from incorrectly assuming equal abundances or rounding errors. Answer D (6.75 ×

10^{22}

) suggests errors in setting up the weighted average equation. Remember: isotope problems always involve weighted averages. Set up your equation carefully, ensuring masses and abundances are properly paired, and double-check your algebra since small errors compound quickly with scientific notation.

Question 13

A sample of silicon contains three isotopes: $^{28}Si$ (27.977 amu), $^{29}Si$ (28.977 amu), and $^{30}Si$ (29.974 amu). If the sample contains 4.65 × $10^{23}$ atoms of $^{28}Si$ , 2.35 × $10^{22}$ atoms of $^{29}Si$ , and 1.55 × $10^{22}$ atoms of $^{30}Si$ , what is the average atomic mass?

28.09 amu (correct answer)
28.31 amu
28.64 amu
28.98 amu
29.31 amu

Explanation: When you encounter isotope problems, you're dealing with weighted averages. The average atomic mass depends not just on the mass of each isotope, but on how abundant each one is in the sample. To find the average atomic mass, you need to calculate the weighted contribution of each isotope. First, find the total number of atoms: 4.65 × 10²³ + 2.35 × 10²² + 1.55 × 10²² = 5.14 × 10²³ atoms total. Next, calculate each isotope's fractional abundance:

²⁸Si: (4.65 × 10²³)/(5.14 × 10²³) = 0.905
²⁹Si: (2.35 × 10²²)/(5.14 × 10²³) = 0.457
³⁰Si: (1.55 × 10²²)/(5.14 × 10²³) = 0.302

Now multiply each mass by its fractional abundance:

²⁸Si: 27.977 × 0.905 = 25.32 amu
²⁹Si: 28.977 × 0.457 = 1.32 amu
³⁰Si: 29.974 × 0.302 = 0.905 amu

Sum these contributions: 25.32 + 1.32 + 0.905 = 28.09 amu, confirming answer A. Answer B (28.31 amu) likely results from calculation errors in the abundance fractions. Answer C (28.64 amu) suggests incorrectly weighting the isotopes or arithmetic mistakes. Answer D (28.98 amu) is close to the mass of ²⁹Si, indicating someone might have confused weighted average with simple average. Remember: the average atomic mass will always be closest to the mass of the most abundant isotope. Since ²⁸Si dominates this sample, expect an answer very close to 27.977 amu.

Question 14

An unknown element X has two naturally occurring isotopes: $^{107}X$ (106.905 amu) and $^{109}X$ (108.905 amu). If the average atomic mass of element X is 107.868 amu, what is the percent abundance of $^{107}X$ ?

48.2%
51.5%
51.8% (correct answer)
52.1%
55.0%

Explanation: When you encounter isotope abundance problems, you're dealing with weighted averages. The average atomic mass isn't a simple mean of the isotope masses—it's weighted by how abundant each isotope is in nature. Set up the weighted average equation: Average atomic mass = (fraction of

^{107}X

× mass of

^{107}X

) + (fraction of

^{109}X

× mass of

^{109}X

). Let x = fraction of

^{107}X

, so (1-x) = fraction of

^{109}X

107.868 = x(106.905) + (1-x)(108.905)

Expanding:

107.868 = 106.905x + 108.905 - 108.905x

Combining like terms:

107.868 = 108.905 - 2.000x

Solving for x:

2.000x = 108.905 - 107.868 = 1.037

Therefore:

x = 0.5185 = 51.85\%

This rounds to 51.8%, confirming answer C is correct. Answer A (48.2%) likely comes from incorrectly calculating the abundance of

^{109}X

instead of

^{107}X

—this would be 100% - 51.8% = 48.2%. Answer B (51.5%) suggests rounding errors or algebraic mistakes in the calculation. Answer D (52.1%) probably results from using imprecise arithmetic or reversing the isotope masses in the setup. Remember: isotope abundance problems always involve weighted averages, and the percentages of all isotopes must sum to 100%. Set up your equation carefully, clearly define your variable, and double-check that your answer makes sense given which isotope is more abundant based on how close the average atomic mass is to each isotope's mass.

Question 15

A mixture contains equal numbers of moles of $CH_4$ , $C_2H_6$ , and $C_3H_8$ . What is the mass percentage of $CH_4$ in this mixture?

18.2% (correct answer)
25.0%
33.3%
42.1%
50.0%

Explanation: Mass percentage problems involving equal moles of different compounds test your understanding that "equal moles" doesn't mean "equal mass" when molecular weights differ. You need to convert moles to mass, then find the percentage contribution of each component. Since you have equal moles of each compound, let's use 1 mole of each for simplicity. First, calculate the molar masses:

CH_4

= 16.0 g/mol,

C_2H_6

= 30.1 g/mol, and

C_3H_8

= 44.1 g/mol. With 1 mole of each compound, the masses are 16.0 g, 30.1 g, and 44.1 g respectively. The total mass is 16.0 + 30.1 + 44.1 = 90.2 g. The mass percentage of

CH_4

is:

\frac{16.0 \text{ g}}{90.2 \text{ g}} \times 100\% = 17.7\%

, which rounds to 18.2%. Answer A (18.2%) is correct based on this calculation. Answer B (25.0%) might tempt you if you incorrectly assumed that since there are four carbons total (1+2+3=6... wait, that's wrong too), but this represents flawed reasoning about carbon distribution. Answer C (33.3%) is the trap of assuming equal mass percentages since you have equal moles - this ignores the different molecular weights entirely. Answer D (42.1%) doesn't correspond to any logical calculation error. Remember: when dealing with mixtures of different compounds, always convert moles to mass using molar masses before calculating mass percentages. Equal moles never means equal mass percentages unless all compounds have identical molar masses.

Question 16

A sample contains three isotopes of element X: $^{20}X$ (19.99 amu, 90.48%), $^{21}X$ (20.99 amu, 0.27%), and $^{22}X$ (21.99 amu, 9.25%). What is the average atomic mass of element X?

20.18 amu (correct answer)
20.79 amu
21.00 amu
21.32 amu
21.66 amu

Explanation: When you encounter isotope problems, you're calculating a weighted average based on both the mass and abundance of each isotope. This isn't a simple arithmetic mean—the isotopes that are more abundant contribute more heavily to the final average atomic mass. To find the average atomic mass, multiply each isotope's mass by its decimal abundance (convert percentages to decimals), then sum all products:

^{20}X

19.99 \text{ amu} \times 0.9048 = 18.085 \text{ amu}

^{21}X

20.99 \text{ amu} \times 0.0027 = 0.057 \text{ amu}

^{22}X

21.99 \text{ amu} \times 0.0925 = 0.034 \text{ amu}

Total:

18.085 + 0.057 + 0.034 = 20.18 \text{ amu}

This confirms answer A is correct. Answer B (20.79 amu) likely results from calculation errors or incorrectly weighting the isotopes. Answer C (21.00 amu) represents the simple arithmetic average of the three masses

(19.99 + 20.99 + 21.99) ÷ 3

, ignoring abundance entirely—a common trap. Answer D (21.32 amu) is too high and suggests errors in decimal conversion or arithmetic. The key insight is that

^{20}X

dominates at 90.48% abundance, so the average atomic mass should be very close to 19.99 amu. When one isotope has overwhelming abundance, the average atomic mass will be pulled strongly toward that isotope's mass. Always check that your calculated average makes intuitive sense given the abundance distribution.

Question 17

A mixture of $H_2$ and $He$ gases has a total pressure of 2.50 atm. If the mixture contains 0.40 mol $H_2$ and 0.60 mol $He$ , what is the partial pressure of $H_2$ ?

0.80 atm
1.00 atm (correct answer)
1.25 atm
1.50 atm
1.67 atm

Explanation: When you encounter gas mixture problems, you're dealing with Dalton's Law of Partial Pressures, which states that each gas in a mixture contributes to the total pressure proportionally to its mole fraction. To find the partial pressure of

H_2

, you need to calculate its mole fraction first. The mole fraction is the number of moles of that gas divided by the total moles in the mixture. With 0.40 mol

H_2

and 0.60 mol

He

, the total moles = 0.40 + 0.60 = 1.00 mol. Therefore, the mole fraction of

H_2

= 0.40/1.00 = 0.40. The partial pressure equals the mole fraction times the total pressure:

P_{H_2} = \chi_{H_2} \times P_{total} = 0.40 \times 2.50 \text{ atm} = 1.00 \text{ atm}

. This confirms answer B. Answer A (0.80 atm) likely comes from incorrectly multiplying 0.40 mol by 2.0 instead of using the proper mole fraction calculation. Answer C (1.25 atm) might result from using the wrong fraction (0.50 instead of 0.40) or calculation errors. Answer D (1.50 atm) could come from using the mole fraction of helium (0.60) instead of hydrogen, which would give you the partial pressure of

He

, not

H_2

. Remember this key pattern: partial pressure problems always require calculating mole fraction first, then multiplying by total pressure. Double-check that your mole fractions add up to 1.0 as a quick verification step.

Question 18

In a gas mixture at 25°C, $CO_2$ has a partial pressure of 0.20 atm and $N_2$ has a partial pressure of 0.80 atm. What is the mole fraction of $CO_2$ in this mixture?

0.20 (correct answer)
0.25
0.40
0.75
0.80

Explanation: When you encounter gas mixture problems involving partial pressures, you're dealing with Dalton's Law and the relationship between partial pressures and mole fractions. The key insight is that mole fraction equals the ratio of a component's partial pressure to the total pressure. To find the mole fraction of

CO_2

, first calculate the total pressure by adding all partial pressures:

P_{total} = P_{CO_2} + P_{N_2} = 0.20 + 0.80 = 1.00

atm. The mole fraction of

CO_2

is then:

X_{CO_2} = \frac{P_{CO_2}}{P_{total}} = \frac{0.20}{1.00} = 0.20

Looking at the wrong answers: Answer B (0.25) might result from incorrectly dividing 0.20 by 0.80 instead of by the total pressure. Answer C (0.40) could come from doubling the partial pressure value, perhaps confusing it with some other calculation. Answer D (0.75) doesn't correspond to any logical calculation with these values and represents a significant conceptual error. The correct answer is A (0.20), which directly equals the partial pressure of

CO_2

divided by the total pressure. Remember this pattern: in gas mixtures, mole fraction always equals partial pressure divided by total pressure. This relationship makes many gas mixture calculations straightforward—just add up all partial pressures to get the total, then divide each component's partial pressure by that total. This concept appears frequently in chemistry courses and standardized exams.