Lesson 9 of 11Quantities in Chemical Reactions

Unit 3 · Lesson 9 of 11

Mass-to-Mass Stoichiometry

Calculating the mass of a reactant or product from the mass of another reactant or product, by chaining mole conversions on either side of a mole-ratio step.

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Key things to remember

1Roadmap: mass given $\to$ mol given ( $\div M_{given}$ ) $\to$ mol wanted (mole ratio) $\to$ mass wanted ( $\times M_{wanted}$ ).
2 $n = m / M$ and $m = n \times M$ are used once on each side of the mole-ratio step.
3Masses of two different substances can never be compared or scaled directly — only their moles are related by the equation's coefficients.
4The chemical equation must be balanced before any of this works.
5Check with conservation of mass: total mass of reactants used = total mass of products formed.
6Keep extra digits in intermediate values; round only the final answer to the sig figs of the given data.

This is where every earlier skill in the unit comes together. In practice, you're almost never given moles directly — you're given a mass you weighed on a balance, and you want to predict a mass of something else. The trick is that a balanced equation only ever talks in moles, so mass has to be converted to moles, carried through the mole ratio, and converted back to mass at the very end. There is no shortcut that skips the middle: masses of two different substances can never be compared or scaled directly, because they generally have different molar masses.

Why Mass Can't Skip the Mole Step

Look at $2 Mg + O_{2} \to 2 MgO$ . The coefficients of Mg and MgO are equal, so a tempting shortcut says " $6.00 g$ of Mg gives $6.00 g$ of MgO". That can't be right: each magnesium atom picks up an oxygen atom, so every mole of product ( $40.31 g$ ) is heavier than the mole of metal it came from ( $24.31 g$ ). Equal moles are not equal masses. The correct answer, $9.95 g$ , only appears once you go through moles. Conservation of mass then gives a free check: the product gained $9.95 - 6.00 = 3.95 g$ , and that is exactly the mass of $O_{2}$ the mole ratio predicts ( $0.1234 mol \times 32.00 g/mol = 3.95 g).$

Fig 1The full mass-to-mass roadmap: mass given → mol given → mol wanted → mass wanted.

The Four-Step Roadmap

Write and balance the chemical equation (if not already given).
Convert the given mass to moles: $n_{given} = \frac{m _{given}}{M _{given}}$ .
Convert to moles of the wanted substance using the mole ratio from the coefficients: $n_{wanted} = n_{given} \times \frac{coeff. wanted}{coeff. given} .$
Convert moles of the wanted substance to mass: $m_{wanted} = n_{wanted} \times M_{wanted} .$

m_{given} \div M_{given} n_{given} \times \frac{coeff. wanted}{coeff. given} n_{wanted} \times M_{wanted} m_{wanted}

Two different molar masses, used at two different points

You'll calculate two separate molar masses in most mass-to-mass problems — one for the given substance (to find its moles) and a different one for the wanted substance (to find its mass). Mixing them up is one of the most common errors in this type of problem.

A tidy layout prevents most errors. Write the balanced equation, then build a small table underneath it with one column per substance and rows for molar mass, moles and mass. Fill in what you know, and each blank cell is one multiplication or division away from a filled one. Label every number with its substance (" $0.2468 mol Mg$ ", not just " $0.2468$ "), keep at least one extra significant figure in intermediate values, and round only the final answer to the number of sig figs in the given data.

Quantity	$2 Mg$	$O_{2}$	$2 MgO$
$M$ (g/mol)	24.31	32.00	40.31
$n$ (mol)	0.2468 (from 6.00 ÷ 24.31)	0.1234 (× 1/2)	0.2468 (× 2/2)
$m$ (g)	6.00 (given)	3.95	9.95

Stoichiometry table for burning 6.00 g of magnesium. The mass row balances: 6.00 g + 3.95 g = 9.95 g.

Working Backward, Scaling Up and Checking

The roadmap works in any direction. "What mass of limestone must be heated to make $5.00 g$ of $CO_{2}$ ?" simply makes the product the given substance and the reactant the wanted one; the steps are identical. It also works at any scale. Industrial chemists work in kilograms and tonnes, and since $\frac{kg}{g/mol}$ gives kilomoles, you can run the whole calculation in kg and kmol without converting to grams. Scaled-up answers can be surprising: burning $1.00 kg$ of propane ( $1000 \div 44.11 = 22.67 mol$ , giving $3 \times 22.67 = 68.01 mol$ of $CO_{2}$ ) produces about $2.99 kg$ of $CO_{2}$ , nearly three times the fuel's mass, because most of that mass is oxygen pulled from the air. Before accepting any answer, ask whether its size makes sense and, when you know all the masses, whether they obey conservation of mass.

Process	Balanced equation	Given	Result
Burning barbecue propane	$C_{3} H_{8} + 5 O_{2} \to 3 CO_{2} + 4 H_{2} O$	1.00 kg propane	2.99 kg $CO_{2}$
Lime kiln (making quicklime for cement)	$CaCO_{3} \to CaO + CO_{2}$	1.00 t limestone	0.560 t CaO and 0.440 t $CO_{2}$
Blast furnace (making iron)	$Fe_{2} O_{3} + 3 CO \to 2 Fe + 3 CO_{2}$	1.00 kg $Fe_{2} O_{3}$	0.699 kg Fe

The same four steps at industrial scale. In the kiln, the two product masses add back up to the 1.00 t of limestone.

AForward (reactant given)

Given: mass of a reactant
Wanted: mass of a product (or of another reactant needed)
Example: $6.00 g Mg \to 9.95 g MgO$

BBackward (product given)

Given: mass of product you need to make
Wanted: mass of reactant to start with
Example: $5.00 g CO_{2} \leftarrow 11.4 g CaCO_{3}$

Recognising the question type

A mass of one substance given and a mass of a different substance asked for, with only one starting amount, is mass-to-mass stoichiometry. If masses of two reactants are given, it becomes a limiting-reagent problem; if an actual measured product mass is also given, finish with percent yield.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

For

2 Mg + O_{2} \to 2 MgO

, what mass of

MgO

forms from

6.00 g

of Mg (

M (Mg) = 24.31

M (MgO) = 40.31 g/mol

Example 2

For

C_{3} H_{8} + 5 O_{2} \to 3 CO_{2} + 4 H_{2} O

, what mass of

CO_{2}

is produced when

22.0 g

C_{3} H_{8}

burns completely (

M (C_{3} H_{8}) = 44.11

M (CO_{2}) = 44.01 g/mol

Example 3

In a blast furnace, iron(III) oxide is reduced by carbon monoxide:

Fe_{2} O_{3} + CO \to Fe + CO_{2}

(unbalanced). What mass of iron can be obtained from

1.00 kg

Fe_{2} O_{3}

Example 4

For

CaCO_{3} \to CaO + CO_{2}

, what mass of

CaCO_{3}

is needed to produce

5.00 g

CO_{2}

(

M (CO_{2}) = 44.01

M (CaCO_{3}) = 100.09 g/mol

Where marks get lost

Trying to scale masses directly by the coefficient ratio, e.g. assuming twice the coefficient means twice the mass (only moles scale that way, not mass, since molar masses differ).
Using the wrong molar mass at the wrong step — e.g. dividing by the wanted substance's molar mass instead of the given substance's.
Stopping at moles of the wanted substance and reporting that number as if it were already the mass answer.
Skipping the balancing step and using coefficients from an unbalanced (or incorrectly balanced) equation.
Rounding intermediate values too early, e.g. rounding $0.2468 mol$ to $0.25 mol$ and getting $10.1 g$ of MgO instead of $9.95 g$ .

Flashcards 9

Four-step mass-to-mass roadmap

Can you scale masses directly using an equation's coefficients?

What must be true of an equation before doing any mass-to-mass stoichiometry?

How many separate molar mass calculations does a typical mass-to-mass problem need?

+ 5 more

Practice

For

2 Na + Cl_{2} \to 2 NaCl

, what mass of NaCl forms from

4.60 g

of Na (

M (Na) = 22.99

M (NaCl) = 58.44 g/mol

Why can't you find the mass of product B directly from the mass of reactant A without converting to moles first?

For

CaCO_{3} \to CaO + CO_{2}

, what mass of

CO_{2}

is produced from

25.0 g

CaCO_{3}

(

M (CaCO_{3}) = 100.09

M (CO_{2}) = 44.01 g/mol

In a mass-to-mass stoichiometry problem, which two molar masses do you need, and why are they different?

Order the correct steps to find the mass of CO2 produced from a given mass of propane burning.

1Convert given mass of propane to moles
2Convert moles of CO2 to mass
3Apply the mole ratio to get moles of CO2
4Balance the combustion equation

For

2 Al + 3 Cl_{2} \to 2 AlCl_{3}

, what mass of

AlCl_{3}

(

M = 133.33 g/mol

) forms from

5.00 g

of Al (

M = 26.98 g/mol

), assuming excess

Cl_{2}

A student calculates

n (CO_{2})

correctly, then reports that value (in mol) as the final mass answer in grams. What step did they skip?

For

N_{2} + 3 H_{2} \to 2 NH_{3}

, what mass of

H_{2}

(

M = 2.02 g/mol

) is required to fully react with

10.0 g

N_{2}

(

M = 28.02 g/mol

Match each stage of a mass-to-mass calculation to the operation it uses.

A.balance the equation
B.divide by $M$ of the given substance
C.multiply by $M$ of the wanted substance
D.multiply by coefficient wanted ÷ coefficient given

mass given → mol given

mol given → mol wanted

mol wanted → mass wanted

before any calculation

Using

Fe_{2} O_{3} + 3 CO \to 2 Fe + 3 CO_{2}

, what mass of iron can be made from

250. g

Fe_{2} O_{3}

(

M = 159.70 g/mol

; Fe = 55.85)? Give 3 sig figs.

Methane burns by

CH_{4} + 2 O_{2} \to CO_{2} + 2 H_{2} O

. What mass of

O_{2}

is needed to burn

10.0 g

CH_{4}

completely? Give 3 sig figs.

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