Converting Between Mass, Moles, and Particles质量、物质的量、粒子数的换算

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 1~35 min

Learning goals

  • Convert between mass, moles, and number of particles using molar mass and Avogadro’s number
  • Explain why every conversion between mass and particles goes through moles
  • Round each answer to the significant figures of the given value

Before you start

  • The mole and Avogadro’s number (2.6.1)
  • Dimensional analysis (2.1.2)

Count molecules by weighing

A bag of sugar is labeled 500 g. How many sugar molecules are inside? You don’t count them; you calculate. Molar mass turns the mass into moles, and Avogadro’s number turns moles into molecules. This lesson makes those two steps automatic, because nearly every calculation in chemistry depends on them.

Moles are the bridge

The last lesson gave two relationships. Molar mass connects mass and moles, and Avogadro’s number connects moles and particles:

n=mMN=n×NAn = \frac{m}{M} \qquad N = n \times N_\text{A}

Put the three quantities in a row, with moles in the middle:

mass m (g) ⇄ ÷ M or × M ⇄ moles n (mol) ⇄ × NA or ÷ NA ⇄ particles N

  • Going right: grams ÷ molar mass = moles; moles × NA = particles.
  • Going left: particles ÷ NA = moles; moles × molar mass = grams.
  • There is no direct step between grams and particles. Always go through moles.

This lesson uses the atomic masses on the AP periodic table (H 1.008, C 12.01, O 16.00, Na 22.99, Cl 35.45, Fe 55.85) and NA = 6.022 × 10²³ mol⁻¹.

Setting it up with conversion factors

US textbooks usually solve these with dimensional analysis (2.1.2): write the molar mass as a conversion factor such as (1 mol H₂O / 18.02 g H₂O) and let the units cancel. Chinese textbooks write formulas instead, as in n = m ÷ M = 36.0 g ÷ 18 g/mol = 2.00 mol, naming each quantity. Both do the same arithmetic. Use whichever you like, but keep the units in every step.

Round the answer to the significant figures of the given value (2.1.1). Molar masses and NA usually have more digits than the data, so they don’t limit the answer.

Pick a substance and a given quantity to see all three values change together. Then press “New practice problem” and try one yourself:

Try it: mass ⇄ moles ⇄ particles
Given:

Type particle counts as 3.01e23 or 3.01×10^23. Results are rounded to the sig figs you enter.

Mass
Moles

Example 1: grams to moles to molecules

How many moles and how many molecules are in 10.0 g of ammonia, NH₃?

Step-by-step solution

  1. Molar mass. Add the atomic masses: NH₃ = 14.01 + 3(1.008) = 17.03 g/mol.

  2. Grams to moles. Divide by the molar mass; g cancels and mol remains:

    n=10.0 g17.03 g/mol=0.5872≈0.587 mol\begin{aligned} n &= \frac{10.0\ \text{g}}{17.03\ \text{g/mol}} \\ &= 0.5872 \approx \boxed{0.587\ \text{mol}} \end{aligned}

  3. Moles to molecules. Use the unrounded 0.5872:

    N=0.5872×6.022×1023=3.54×1023\begin{aligned} N &= 0.5872 \times 6.022 \times 10^{23} \\ &= \boxed{3.54 \times 10^{23}} \end{aligned}

  4. Sig figs. 10.0 g has three significant figures, so both answers get three.

  5. The road map. Grams ⇄ moles ⇄ particles: moles are always the middle stop.

Example 2: atoms to grams

What is the mass of 2.50 × 10²² atoms of iron?

Step-by-step solution

  1. Plan. Atoms → moles → grams. There is no shortcut from atoms straight to grams.

  2. Atoms to moles.

    n=2.50×10226.022×1023=0.04151 mol\begin{aligned} n &= \frac{2.50 \times 10^{22}}{6.022 \times 10^{23}} \\ &= 0.04151\ \text{mol} \end{aligned}

  3. Moles to grams. M(Fe) = 55.85 g/mol:

    m=0.04151×55.85=2.32 g\begin{aligned} m &= 0.04151 \times 55.85 \\ &= \boxed{2.32\ \text{g}} \end{aligned}

  4. Check. 2.50 × 10²² is about 1/24 of a mole, and 55.85 g ÷ 24 ≈ 2.3 g, so 2.32 g is reasonable.

Practice: which conversion gets you to the quantity asked for?
  1. Given grams, find moles
  2. Given moles, find grams
  3. Given moles, find molecules
  4. Given a number of atoms, find moles
  5. Given grams, find molecules
  6. Given a number of formula units, find grams
  7. A 2.8-g piece of iron: how many Fe atoms?
  8. 1.0 × 10²² formula units of NaCl: what mass?

Common mistakes

Mistake 1: “Grams × NA = particles.” 18 g × 6.022 × 10²³ means nothing, because grams are not moles. Divide by the molar mass first. The units warn you: g × (particles/mol) does not cancel to particles.

Mistake 2: “10.0 g of ammonia contains 3.54 × 10²³ atoms.” 3.54 × 10²³ is the number of molecules. Each NH₃ molecule has 4 atoms, so there are 4 × 3.54 × 10²³ = 1.41 × 10²⁴ atoms. Read whether the question asks for atoms or molecules.

Mistake 3: “The molar mass of oxygen gas is 16.00 g/mol.” Oxygen gas is O₂, 32.00 g/mol; 16.00 g/mol is for O atoms. The same goes for H₂ (2.016 g/mol), N₂ (28.02 g/mol), and Cl₂ (70.90 g/mol). All seven diatomic elements (H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂) are worth knowing by heart.

Quick check

How many moles are in 88.0 g of CO₂?

Which 1.0-g sample contains the most molecules?

Summary

  • Every conversion between grams and particles goes through moles: n = m ÷ M and N = n × NA.
  • To count atoms, multiply the number of molecules by the number of atoms in each molecule.
  • Write each step as a conversion factor with units, keep extra digits in the calculator, and round the final answer to the data’s significant figures.
On to AP (for later)

AP questions often chain these steps, for example: “A tablet contains 0.250 g of aspirin, C₉H₈O₄. How many carbon atoms does it contain?” AP scoring rewards visible work, so write each conversion factor with units and finish with the right number of significant figures.

Practice this chapter: 2.6 · The Mole and Stoichiometry · 10 questions

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