# How to calculate Mass to Mass Stoichiometry

Stoichiometry calculations is grouped into different categories and one of them is mass to mass stoichiometry.

In mass-to-mass stoichiometry, the analysis and comparison are still based on mole concept.

To proceed in the calculations of mass-to-mass stoichiometry, there is a need to recall how to change or convert mole to mass as discussed fully in ‘what is mole in stoichiometry

Recall that one mole of a substance is equal to molar mass of the substance, so this is the basis of the calculation.

I’ll solve some examples of these mass-to-mass stoichiometry problems just to help us understand the concept better.

## 5 Examples on Mass-to-Mass Stoichiometry

1. What is the mass of carbon iv oxide evolved  when 14 grams of carbon burns completely in oxygen  according to this equation

C + O2===CO2

Solution

Step 1

First, you recall that in every stoichiometry you’re comparing two substances.

Step 2

So in this one, you’re comparing carbon and carbon iv oxide.

Step 3

Then, underline the two the number if moles (coefficients) of the two substances in the equation.

Step 4

Then solve the question

So let’s solve like this

Compare

1 mole of carbon ==2 moles of carbon iv oxide

Converting moles to mass

12 g of carbon ==44 grams of CO2

14g of carbon == X

X= 14 X 44/ 12 = 51.33 grams of Carbon iv oxide

2. Consider the equation

Mg + ZnCl2===MgCl2 + Zn

What mass of magnesium chloride is produced when 25 grams of magnesium reacts with

Zinc chloride?

Solution

Step 1

First, you recall that in every stoichiometry you’re comparing two substances.

Step 2

So in this one, you’re comparing Magnesium and Magnesium chloride

Step 3

Then, underline the two the number if moles (coefficients) of the two substances in the equation.

Step 4

Then solve the question

So let’s solve like this

Compare

1 mole of Mg === 1 mole of MgCl2

24 grams of Mg === 95 grams of MgCl2

25 grams of Mg ==== x

X = 25 x 95 / 24

X = 98.96 grams of Magnesium Chloride (MgCl2)

3. Calculate the mass of ammonia produced by reacting 33g of Nitrogen.

N2 + 3H2===2NH3

Solution

Step 1

First, you compare two substances

Step 2

So here we compare two substances Nitrogen and ammonia

Step 3

Then, underline the two the number if moles (coefficients) of the two substances in the equation.

Step 4

Then solve the question

So let’s solve like this

Compare

1 mole of N2==2 moles of NH3

28g of N2 ===34g of NH3

33g of N2=== X

X= 33 X 34/28

× = 40.07 grams of Ammonia (NH3)

4. What mass of carbon iv oxide is produced by completely burning 30g of methane in oxygen

CH4+ 2O2 ==CO2 + 2H2O

Solution

Step 1

First, you recall that in every stoichiometry you’re comparing two substances.

Step 2

So in this one, you’re comparing methane and carbon iv oxide.

Step 3

Then, underline the two the number if moles (coefficients) of the two substances in the equation.

Step 4

Then solve the question

So let’s solve like this;

1 mole of methane ==== 1 mole of carbon iv oxide

16 grams of Methane ==== 44 grams of Carbon iv oxide

30 grams of Methane ==== x

X = 30 x 44 / 16

X = 82.5 grams of carbon iv oxide

5. What mass of sodium is needed to produce 19g of sodium hydroxide on reaction with cold water

2Na + 2H2O===2NaOH +H2

Solution

Step 1

First, you recall that in every stoichiometry you’re comparing two substances.

Step 2

So in this one, you’re comparing carbon and carbon iv oxide.

Step 3

Then, underline the two the number if moles (coefficients) of the two substances in the equation.

Step 4

Then solve the question

So let’s solve like this

Compare the two substances

2 moles of Na ==== 2 moles of NaOH

46 grams of Na === 80 grams of NaOH

X        ==== 19 grams of NaOH

X = 46 X 19 /80

X = 10.93 grams of sodium