Composition of mixtures — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Mass percent composition, percent purity calculation, average atomic mass for isotopic mixtures, interpretation of mass spectrometry data for mixture composition, and applying proportional reasoning to find unknown mixture components.

You should already know: Isotopic notation and basic atomic structure, Mole-mass unit conversions, The law of conservation of mass.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Composition of mixtures?

Composition of mixtures refers to the proportional breakdown of the different components (elements, compounds, or isotopes) that make up a macroscopic mixture. For AP Chemistry, this topic falls within Unit 1: Atomic Structure and Properties; per the 2020 AP Chemistry CED, it aligns with learning objective SAP-2.A, accounting for approximately 8% of the unit exam weight, which translates to 2-4% of the total AP exam score. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections: it is commonly tested as a standalone MCQ, or as the opening calculation step in a longer FRQ involving stoichiometry or atomic structure. Mixtures are classified as homogeneous (uniform composition throughout, e.g., an elemental sample with multiple isotopes, brass alloy) or heterogeneous (non-uniform composition), and AP Chemistry almost exclusively assesses composition of homogeneous mixtures. Synonyms for composition include mass percentage, fractional abundance, and component proportion. The core skill tested here is the ability to calculate unknown component amounts from aggregate experimental data, a foundational skill for all quantitative chemistry that follows.

2. Mass Percent Composition

Mass percent composition (or mass percentage) of a component in a mixture is the percentage of the total mixture mass contributed by that component. It is used for both elemental mixtures (e.g., alloys) and compound mixtures (e.g., a mixture of two salts), and is the standard way to report percent purity of an impure sample (a common AP exam scenario).

The formula for mass percent of component X is: $$\text{Mass percent of X}=\frac{m_X}{m_{\text{total mixture}}}\times 100\%$$ Where $m_X$ is the mass of pure component X, and $m_{\text{total mixture}}$ is the total mass of the full mixture. Intuition: this is just a proportion scaled to 100, so if you have 100 g of mixture, the mass percent is exactly how many grams of X are present. The formula can be rearranged to solve for any unknown: if you know the total mass and mass percent, you can find $m_X=\frac{\text{Mass percent}\times m_{\text{total}}}{100\%}$. For mixtures where all components are accounted for, the sum of all mass percentages will always equal 100%.

Worked Example

A 15.2 g impure sample of sodium chloride (table salt) is purified by recrystallization, yielding 12.8 g of pure NaCl. What is the percent impurity of the original impure sample?

1. First, identify what we need: percent of impurity, so first find the mass of impurity in the original sample.
2. Mass of impurity = total mass of original sample - mass of pure NaCl = $15.2\text{g}-12.8\text{g}=2.4\text{g}$.
3. Plug into the mass percent formula: $\text{Mass percent impurity}=\frac{2.4\text{g}}{15.2\text{g}}\times 100\%=15.8\%$.
4. Check: percent purity of NaCl is $\frac{12.8}{15.2}\times 100\%=84.2\%$, and $15.8\%+84.2\%=100\%$, which matches the requirement for total composition.

Exam tip: Always underline which component the question asks for (impurity vs pure compound) before starting calculations. It is extremely common for students to report the wrong percentage on exam questions.

3. Average Atomic Mass of Isotopic Mixtures

All naturally occurring elements are homogeneous mixtures of isotopes: atoms of the same element with different masses due to differing numbers of neutrons. The average atomic mass reported on the periodic table is a weighted average of the masses of each stable isotope, weighted by their fractional abundances in the natural mixture.

Fractional abundance is the proportion (as a decimal, not a percentage) of each isotope in the mixture, so the sum of all fractional abundances for an element will always equal 1. The formula for average atomic mass ($M_{avg}$) is: $$M_{avg}=\sum (f_i\times M_i)$$ Where $f_i$ = fractional abundance of isotope $i$, and $M_i$ = isotopic mass of isotope $i$. Intuition: the more abundant an isotope is, the more it pulls the average mass toward its own mass, so the average atomic mass will always be closer to the mass of the most abundant isotope. This is one of the most frequently tested skills in the composition of mixtures topic on the AP exam.

Worked Example

Boron has two stable isotopes: ¹⁰B (mass = 10.013 amu, 19.9% abundance) and ¹¹B (mass = 11.009 amu, 80.1% abundance). Calculate the average atomic mass of naturally occurring boron.

1. Convert percentage abundances to fractional abundances by dividing by 100%: $f_{{}^{10}B}=0.199$, $f_{{}^{11}B}=0.801$.
2. Confirm fractional abundances sum to 1: $0.199+0.801=1.000$, so all isotopes are accounted for.
3. Calculate the weighted sum: $M_{avg}=(0.199\times 10.013)+(0.801\times 11.009)$.
4. Compute and round: $1.993+8.818=10.811$ amu, which rounds to 10.81 amu, matching the periodic table value for boron.

Exam tip: Always convert percentages to decimals before plugging into the formula. If you use percentages directly, you will get an answer 100x too large, which will be marked incorrect even if your arithmetic is right.

4. Mixture Composition from Mass Spectrometry

Mass spectrometry is an experimental technique that separates charged particles by their mass-to-charge ($m/z$) ratio, producing a spectrum with peaks where the x-axis is $m/z$ (equal to the mass of the particle for +1 charge) and the y-axis (peak area or height) is proportional to the relative abundance of that component. This makes mass spectrometry the primary experimental method for determining the composition of isotopic mixtures, and can also be used to find the composition of mixtures of different compounds.

To get fractional abundances from a mass spectrum: the abundance of a component is proportional to its peak intensity, so you sum the intensities of all peaks to get the total intensity, then divide each individual peak intensity by the total to get fractional abundance. This works for any mixture separated by mass spectrometry, from isotopes to organic compounds.

Worked Example

The mass spectrum of a sample of argon shows three peaks with the following relative intensities: $m/z=36$ (intensity = 0.337), $m/z=38$ (intensity = 0.063), $m/z=40$ (intensity = 99.600). What is the fractional abundance of ⁴⁰Ar?

1. Calculate the total intensity of all peaks by summing individual intensities: $0.337+0.063+99.600=100.000$.
2. Fractional abundance of ⁴⁰Ar is the intensity of its peak divided by total intensity: $f_{{}^{40}Ar}=\frac{99.600}{100.000}=0.996$.
3. Confirm by calculating the other abundances: $f_{{}^{36}Ar}=0.00337$, $f_{{}^{38}Ar}=0.00063$, sum = $0.996+0.00337+0.00063=1.000$, so the calculation is correct.

Exam tip: Always sum all peak intensities explicitly, even if they look like they will add to 100. Small measurement errors or unlabeled minor peaks can shift the total, leading to incorrect abundance values.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using percentage abundances instead of fractional abundances in the average atomic mass formula, getting a 2400 amu average for magnesium instead of 24 amu. Why: Students confuse percentage and decimal abundance, and skip the conversion step. Correct move: Always write the conversion step explicitly, and check that your final average falls between the lowest and highest isotopic masses.
• Wrong move: When calculating mass percent, using the mass of one component as the total mass instead of the mass of the full mixture. Why: Students misread the problem and misidentify the denominator. Correct move: Explicitly label $m_{\text{total}}$ as the mass of the full mixture/impure sample before starting any calculation.
• Wrong move: Taking the simple unweighted average of isotopic masses instead of the weighted average. Why: Students assume equal abundance for all isotopes by default. Correct move: Always multiply each isotopic mass by its abundance before summing; never add and divide by the number of isotopes.
• Wrong move: Using m/z values instead of peak intensities to calculate abundances in mass spectrometry. Why: Students confuse the mass of a component with how much of it is present. Correct move: Remember: m/z gives mass, peak intensity gives abundance; always use intensity for abundance calculations.
• Wrong move: Reporting percent purity when the question asks for percent impurity. Why: Students misread the prompt and stop at the first calculation. Correct move: Underline the requested quantity before starting, and confirm you answered the question asked before moving on.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

A 4.00 g impure sample of silver oxide (Ag₂O) is decomposed, yielding 3.12 g of pure silver metal. Assuming silver is only present in Ag₂O, what is the mass percent of pure Ag₂O in the original sample? (Molar mass of Ag = 107.87 g/mol, molar mass of Ag₂O = 231.74 g/mol) A) 78.0% B) 83.8% C) 90.0% D) 128%

Worked Solution: First calculate the mass fraction of Ag in pure Ag₂O: $\frac{2(107.87)}{231.74}=\frac{215.74}{231.74}\approx 0.931$. The mass of pure Ag₂O is equal to the mass of Ag divided by this mass fraction: $\frac{3.12\text{g}}{0.931}\approx 3.35\text{g}$ pure Ag₂O. The mass percent is $\frac{3.35\text{g}}{4.00\text{g}}\times 100\%\approx 83.8\%$. Option A is the mass percent of silver in the original sample, option C is a distractor from incorrect molar mass calculation. The correct answer is B.

Question 2 (Free Response)

Thallium has two stable isotopes: ²⁰³Tl and ²⁰⁵Tl. The average atomic mass of thallium is 204.38 amu. (a) Calculate the fractional abundance of each isotope. (b) Explain why the average atomic mass is closer to 205 amu than 203 amu. (c) Thallium forms a compound TlCl with chlorine. Chlorine has two isotopes: ³⁵Cl (fractional abundance 0.75) and ³⁷Cl (0.25). How many distinct $m/z$ peaks (for +1 charge on the full TlCl molecule) will appear in the mass spectrum? List their masses.

Worked Solution: (a) Let $f_{203}=x$, so $f_{205}=1-x$. Substitute into the average atomic mass formula: $$204.38=203x+205(1-x)$$ Expand and rearrange: $204.38=205-2x⟹2x=205-204.38=0.62⟹x=0.31$. Thus $f_{{}^{203}Tl}=0.31$, $f_{{}^{205}Tl}=0.69$. Check: $0.31(203)+0.69(205)=62.93+141.45=204.38$ amu, which matches the given average.

(b) Average atomic mass is a weighted average, so it is pulled toward the mass of the more abundant isotope. ²⁰⁵Tl has a higher fractional abundance (0.69) than ²⁰³Tl (0.31), so the average is closer to 205 amu.

(c) There are 4 possible mass combinations, two of which are unique:

1. ²⁰³Tl + ³⁵Cl = 203 + 35 = 238
2. ²⁰³Tl + ³⁷Cl = 203 + 37 = 240
3. ²⁰⁵Tl + ³⁵Cl = 205 + 35 = 240
4. ²⁰⁵Tl + ³⁷Cl = 205 + 37 = 242 The distinct peaks are at 238, 240, and 242, so 3 distinct peaks.

Question 3 (Application / Real-World Style)

14-karat gold is a gold alloy mixture used for jewelry, required to be 58.3% gold by mass to meet legal standards. A 12.0 g ring claimed to be 14-karat gold is analyzed, and found to contain 6.7 g of pure gold. Does this ring meet the legal standard for 14-karat gold? Justify your answer with calculations.

Worked Solution: First calculate the mass percent of gold in the ring: $\frac{\text{mass of gold}}{\text{total mass of ring}}\times 100\%=\frac{6.7\text{g}}{12.0\text{g}}\times 100\%\approx 55.8\%$ gold by mass. The legal standard for 14-karat gold is 58.3% gold by mass. 55.8% is less than the required 58.3%, so this ring does not meet the legal standard for 14-karat gold; it has a lower gold content than required.

7. Quick Reference Cheatsheet

8. What's Next

Composition of mixtures is a foundational quantitative skill that is required for almost every other unit in AP Chemistry. Next, you will apply the composition skills you learned here to empirical and molecular formula calculations, which rely on mass percent data to find the formula of an unknown compound. Without mastering mixture composition, stoichiometric calculations involving impure reactants (a common FRQ scenario) will be impossible, as you will not be able to find the actual mass of reactive compound present. This topic also underpins the study of solution concentration in Unit 3, where mass percent, mole fraction, and other concentration units are just different ways to express the composition of a solute-solvent mixture. The core skill of weighted averages for isotopic mixtures also prepares you for acid-base and equilibrium calculations later in the course.

• Empirical and molecular formulas
• Mass spectrometry of elements
• Solution concentration
• Stoichiometry of impure reactants