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
- 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