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How do we convert between the mass of a substance and the number of particles or moles it contains?

Calculate molar mass, convert between mass, moles, and particles, and find percent composition and empirical formulas (MA STE HS-PS1-7(MA), proportional reasoning with chemical formulas).

A standard-level answer on molar mass and percent composition for Massachusetts high school chemistry: finding molar mass from a formula, converting between mass, moles, and particles with Avogadro's number, and calculating percent composition and empirical formulas, grounded in HS-PS1-7(MA).

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  1. What this topic is asking
  2. Molar mass
  3. Converting between mass, moles, and particles
  4. Percent composition
  5. Empirical formulas from percent composition
  6. Try this

What this topic is asking

Standard HS-PS1-7(MA) expects you to use proportional reasoning with chemical formulas to solve quantitative problems. The unit that makes this possible is the mole, introduced in Module 1. This page turns the mole into a working tool: finding the molar mass of a substance, converting between mass, moles, and particles, and using a formula to find percent composition and an empirical formula.

Molar mass

To find molar mass, add the atomic mass of every atom in the formula. For water, H2O\text{H}_2\text{O}: 2(1)+16=182(1) + 16 = 18 g/mol. For calcium carbonate, CaCO3\text{CaCO}_3: 40+12+3(16)=10040 + 12 + 3(16) = 100 g/mol. The atomic masses come straight from the periodic table; the connection between atomic mass and the mole is set out in average atomic mass and the mole concept.

Converting between mass, moles, and particles

The mole sits at the center of two conversions:

  • Mass and moles use molar mass: n=mMn = \dfrac{m}{M}, rearranged to m=nMm = nM.
  • Moles and particles use Avogadro's number: one mole contains 6.022×10236.022 \times 10^{23} particles, so number of particles =n×6.022×1023= n \times 6.022 \times 10^{23}.

Think of moles as the hub: mass converts to moles using MM, and moles convert to particles using NAN_A. To go from mass straight to particles, pass through moles.

Percent composition

For carbon dioxide, CO2\text{CO}_2 (molar mass 44 g/mol): carbon is 1244×100=27.3%\dfrac{12}{44} \times 100 = 27.3\% and oxygen is 3244×100=72.7%\dfrac{32}{44} \times 100 = 72.7\%. The percentages always sum to 100 (within rounding), which is a quick check.

Empirical formulas from percent composition

The empirical formula is the simplest whole-number ratio of atoms. Working backwards from percent composition:

  1. Assume a 100 g sample, so each percent becomes that many grams.
  2. Convert each mass to moles using the element's atomic mass.
  3. Divide every mole value by the smallest to get the ratio.
  4. If the ratio is not whole, multiply all values by a small integer to clear it.

Try this

Q1. Find the molar mass of sodium sulfate, Na2SO4\text{Na}_2\text{SO}_4. Use Na = 23, S = 32, O = 16. [1]

  • Cue. 2(23)+32+4(16)=46+32+64=1422(23) + 32 + 4(16) = 46 + 32 + 64 = 142 g/mol.

Q2. How many moles are in 11 g of carbon dioxide (M=44M = 44 g/mol)? [1]

  • Cue. n=1144=0.25n = \dfrac{11}{44} = 0.25 mol.

Exam-style practice questions

Practice questions written in the style of MA DESE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

MA Chemistry (style)3 marksFor glucose C6H12O6\text{C}_6\text{H}_{12}\text{O}_6: (a) Calculate the molar mass. (b) Find the number of moles in 90 g. (c) Find the percent by mass of carbon. Use C = 12, H = 1, O = 16 g/mol.
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A 3-point quantitative item.

(a) 1 point: 6(12)+12(1)+6(16)=72+12+96=1806(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g/mol.
(b) 1 point: n=mM=90180=0.50n = \dfrac{m}{M} = \dfrac{90}{180} = 0.50 mol.
(c) 1 point: percent carbon =72180×100=40%= \dfrac{72}{180} \times 100 = 40\%. Markers reward the correct molar mass first, then both proportional steps.

MA Chemistry (style)2 marksA compound is 40.0% sulfur and 60.0% oxygen by mass. Find its empirical formula. Use S = 32, O = 16 g/mol.
Show worked answer →

A 2-point empirical-formula item.

1 point: assume 100 g, so 40.0 g S and 60.0 g O; moles are 40.0/32=1.2540.0/32 = 1.25 mol S and 60.0/16=3.7560.0/16 = 3.75 mol O.
1 point: divide by the smaller (1.25): S is 1, O is 3, giving SO3\text{SO}_3. Markers reward dividing by the smallest mole value to get the whole-number ratio.

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