# 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 + O_{2}===CO_{2}

*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 CO _{2}*

*14g of carbon == X*

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

2. *Consider the equation*

*Mg + ZnCl _{2}===MgCl_{2} + 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 MgCl _{2}*

*24 grams of Mg === 95 grams of MgCl _{2}*

*25 grams of Mg ==== x*

*X = 25 x 95 / 24*

*X = 98.96 grams of Magnesium Chloride (MgCl _{2})*

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

N_{2} + 3H_{2}===2NH_{3}

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 N_{2}==2 moles of NH_{3}

28g of N_{2 }===34g of NH_{3}

33g of N_{2}=== X

X= 33 X 34/28

× = 40.07 grams of Ammonia (NH_{3})

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

CH_{4}+ 2O_{2} ==CO_{2} + 2H_{2}O

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 + 2H_{2}O===2NaOH +H_{2}

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*