Unit 1 · Topic 1.3 Beta

Elemental Composition of Pure Substances

A pure compound has a fixed composition by mass.

Practice 1: Models and RepresentationsPractice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

A food label says a sweetener is 40% carbon by mass. A forensic lab burns a white powder and weighs the products. In both cases, chemists use mass percents to work out what a substance is made of, atom by atom, and identify it with nothing more than a balance and a periodic table.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. How many moles of oxygen atoms are in 8.00 g of oxygen atoms (16.00 g/mol)?

  1. 0.500 mol
  2. 128 mol
  3. 2.00 mol
  4. 0.0500 mol
Show the answer

8.00 g ÷ 16.00 g/mol = 0.500 mol.

  • Correct: 0.500 mol:
  • 128 mol:
  • 2.00 mol:
  • 0.0500 mol:

2. What is the molar mass of CO₂ (C 12.01, O 16.00)?

  1. 44.01 g/mol
  2. 28.01 g/mol
  3. 60.01 g/mol
  4. 40.01 g/mol
Show the answer

12.01 + 2(16.00) = 44.01 g/mol.

  • Correct: 44.01 g/mol:
  • 28.01 g/mol:
  • 60.01 g/mol:
  • 40.01 g/mol:

Part 4 · See it

See it first

Mass percents (40.00% C, 6.71% H, 53.29% O) become grams in 100 g, then moles (3.331, 6.657, 3.331), then the ratio 1 : 2 : 1, CH₂O; a molar mass of 180.2 g/mol gives six units, C₆H₁₂O₆.
From mass percents to a formula: grams, then moles, then the smallest whole-number ratio, then the molar mass to scale up. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A pure compound is always made of the same atoms in the same ratioit always has the same mass percent of each element (law of definite proportions)
  2. Atoms of different elements have different massesa mass ratio is not an atom ratio, so masses must be converted to moles
  3. Dividing every mole amount by the smallest gives the simplest ratiothat ratio, cleared to whole numbers, is the empirical formula
  4. Many molecules share one empirical formulathe molar mass is needed to find how many empirical units make one molecule

Part 6 · Key ideas

Key ideas

  • A pure substance has a fixed composition (law of definite proportions): every sample of water is 11.2% H by mass.
  • Mass percent of an element = (mass of the element in one formula ÷ molar mass) × 100.
  • Empirical formula: assume 100 g, convert grams to moles, divide by the smallest, clear fractions (×2 for .5, ×3 for .33).
  • Molecular formula = empirical formula × n, where n = molar mass ÷ empirical-formula mass.

Part 7 · Misconception

A common mistake

The wrong idea: Mass percents give the atom ratio directly, so a compound that is 75% X and 25% Y has three X atoms for each Y.

What actually happens: Different atoms have different masses, so a mass ratio is not an atom ratio. Convert each mass to moles first; only the ratio of moles gives the formula.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Burning magnesium in three labs

Three lab groups heat different masses of pure magnesium in air until it has all turned into a white solid, magnesium oxide, and weigh the product.

Mass of magnesium used and of magnesium oxide made
GroupMass of Mg (g)Mass of magnesium oxide (g)Mass of O gained (g)
11.2162.0160.800
23.0405.0402.000
30.72931.20930.4800

1. What is the mass percent of magnesium in Group 1's product?

Type a number in %.

Show the answer

Mass percent Mg = (1.216 g ÷ 2.016 g) × 100 = 60.32%.

  • Answer: 60.32 %

2. Which conclusion do the three groups' results support?

  1. Magnesium oxide is about 60.3% magnesium by mass in each sample
  2. Larger samples of magnesium oxide contain a larger percent of magnesium
  3. Magnesium and oxygen combine in a 1 : 1 ratio by mass
  4. The groups made different compounds, since their product masses differ
Show the answer

All three products are 60.3% Mg by mass. A compound has the same composition by mass whatever the sample size.

  • Correct: Magnesium oxide is about 60.3% magnesium by mass in each sample: Right: 60.32%, 60.32% and 60.31%: a fixed composition, the law of definite proportions.
  • Larger samples of magnesium oxide contain a larger percent of magnesium: The masses differ, but the percent is the same in every group.
  • Magnesium and oxygen combine in a 1 : 1 ratio by mass: Group 1 used 1.216 g Mg with 0.800 g O, a mass ratio of about 1.5 : 1, not 1 : 1.
  • The groups made different compounds, since their product masses differ: Different amounts of one compound have different masses but the same composition.

3. Using Group 2's data, what is the empirical formula of magnesium oxide?

  1. Mg₃O₂
  2. MgO₂
  3. Mg₂O
  4. MgO
Show the answer

Formulas count atoms, so compare moles: 0.1251 mol Mg : 0.1250 mol O = 1 : 1, so MgO.

  • Mg₃O₂: This uses the mass ratio, about 3 : 2, as if it were a ratio of atoms. Convert masses to moles first.
  • MgO₂: This would need twice as many moles of O as Mg; the moles are equal.
  • Mg₂O: This would need twice as many moles of Mg as O; the moles are equal.
  • Correct: MgO: Right: 3.040/24.31 = 0.1251 mol Mg and 2.000/16.00 = 0.1250 mol O, a 1 : 1 ratio.

4. Based on the groups' results, what mass of magnesium is in 12.5 g of magnesium oxide?

Type a number and its unit.

Show the answer

Magnesium oxide is 60.32% Mg by mass: 12.5 g × 0.6032 = 7.54 g of Mg.

  • Answer: 7.54 g

Data table

Analysis of a sweet-tasting compound

A lab analyzes a pure white solid that contains only carbon, hydrogen and oxygen. Its molar mass is found separately to be 180.2 g/mol.

Mass percent of each element
ElementMass percent (%)Atomic mass (amu)
C40.0012.01
H6.711.008
O53.2916.00

5. What is the empirical formula of the compound?

  1. C₄₀H₇O₅₃
  2. C₃H₇O₃
  3. C₆H₁₂O₆
  4. CH₂O
Show the answer

Assume 100 g: 40.00/12.01 = 3.331 mol C, 6.71/1.008 = 6.657 mol H, 53.29/16.00 = 3.331 mol O. Divide by 3.331: 1 : 1.999 : 1, so CH₂O.

  • C₄₀H₇O₅₃: Percents are masses, not atom counts. Convert each to moles first.
  • C₃H₇O₃: This rounds the moles in 100 g (3.33, 6.66, 3.33) instead of dividing by the smallest.
  • C₆H₁₂O₆: That is the molecular formula; the empirical formula is the simplest ratio.
  • Correct: CH₂O: Right: per 100 g, 3.331 mol C, 6.657 mol H, 3.331 mol O; dividing by 3.331 gives 1 : 2 : 1.

6. What is the molar mass of the empirical formula CH₂O (one C, two H and one O)?

Type a number and its unit.

Show the answer

12.01 + 2(1.008) + 16.00 = 30.03 g/mol.

  • Answer: 30.03 g/mol

7. What is the molecular formula of the compound?

  1. CH₂O
  2. C₁₂H₂₄O₁₂
  3. C₆H₁₂O₆
  4. C₃H₆O₃
Show the answer

n = molar mass ÷ empirical-formula mass = 180.2 / 30.03 = 6. Multiply every subscript by 6: C₆H₁₂O₆.

  • CH₂O: The empirical formula has a molar mass of 30.03, not 180.2.
  • C₁₂H₂₄O₁₂: That is twelve CH₂O units, 360.3 g/mol, twice the molar mass measured.
  • Correct: C₆H₁₂O₆: Right: 180.2 ÷ 30.03 = 6.00, so the molecule is six CH₂O units.
  • C₃H₆O₃: This is three CH₂O units, with a molar mass of 90.08 g/mol.

Part 9 · Summary

Summary

A pure compound has a fixed composition by mass. Mass percent is an element's share of the molar mass. To find an empirical formula, convert mass data to moles and reduce to the smallest whole-number ratio. The molar mass then gives the molecular formula, a whole-number multiple of the empirical one.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections