# Mass-to-Mass Stoichiometric Calculations

Learn how to set up and make mole to mass, mass to mole and mass to mass stoichiometric calculations. Learn how the ratios of moles helps you compare and make calculations. Learn how to relate mole ratios to molar mass.

## Mole to Mass Calculations

You have previously learned about mole ratios and how to make mole to mole calculations from balanced chemical equations. This lesson will teach you how to relate the mole ratio to the amounts of substances in grams. For either reactants or products, you will learn how to determine mass when given moles, moles when given mass and mass when given mass.

To start, you know if you have an equation such as 2H2 + O2 –> 2 H2 O, your mole ratios are:

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• 2 moles H2 per 1 mole O2
• 2 moles H2 per 2 moles H2 O
• 1 mole O2 per 2 moles H2 O

If you want to convert moles to mass, you have to do the following conversions. First, convert moles of substance A into moles of substance B. Then, convert moles of B into mass of B. Moles A –> moles B –> mass B.

The main equation is: moles A x (mole ratio of B/A) x molar mass of B = mass of B. The way you would use this in an actual problem is the following. You are asked how many grams of glucose are produced when 3 moles of water react with carbon dioxide. Start with your balanced equation: 6 CO2 + 6 H2 O–> C6 H12 O6 + 6 O2.

Let water equal A, and let glucose equal B. From the balanced reaction, you know the mole ratio of B/A is (1 mole C6 H12 O6 )/(6 moles H2 O). So, in the original equation, you have both of these terms. Moles A x (mole ratio of B/A) x molar mass of B = mass of B.

• Moles A = 3 moles H2 O
• Mole ratio of B/A = (1 mole C6 H12 O6 )/(6 moles H2 O)

You have learned previously how to determine the amount of grams in one mole of a substance. For this problem, I will just tell you it is 180g glucose per 1 mole glucose. Now you have all three pieces of data you need to figure out the solution.

• Moles A x (mole ratio of B/A) x molar mass of B = mass of B
• (3 moles H2 O)/1 x (1 mole C6 H12 O6 )/(6 moles H2 O) x (180g C6 H12 O6 )/(1 mole C6 H12 O6 ) = grams of C6 H12 O6

The ‘moles H2 O’ and ‘mole C6 H12 O6 ‘ cross out: (3/1) x (1/6) x (180g/1) = grams of C6 H12 O6 .

90g = grams of C6 H12 O6 !

## Mass to Mole Calculations

Let’s try another example, this time converting mass to moles. The conversions you need to make are:

• Mass A –> moles A –> moles B
• Mass A x (1 mole A)/(molar mass of A) x (mole ratio of B)/A = moles B

Start with the balanced equation of 5C + 2 SO2 –> CS2 + 4CO. If 8g of SO2 how many moles of CS2 are formed?

In this example, the SO2 A and the CS2 represents B. Instead of you having to figure out the grams of SO2 in one mole, I will tell you that it is 64.1g SO2 /1 mole SO2 %.

Set up the equation:

• (Mass A)/(molar mass of A) x (mole ratio of B)/A = moles B
• (8g SO2 )/1 x (1 mole SO2 )/(64.1g SO2 ) x (1 mole CS2 )/(2 mole SO2 ) = moles CS2 formed

What crosses out? ‘g SO2 ‘ and ‘moles SO2 .’ The equation becomes (8/1) x (1/64.1g) x (1 mol CS2 )/2 = moles CS2 formed.

0.06 moles CS2 = moles CS2 formed!

## Mass to Mass Calculations

Let’s do one last type of calculation, mass to mass. The conversions you need to make are:

• Mass A –> moles A –> moles B –> mass B
• (Mass A)/(molar mass A) x (moles A)/1 x (mole ration B)/A x 1/(moles B) * (molar mass B)/(moles B) = mass of B

How many grams of NH4 NO3 are needed to produce 33g of N2 O in the following reaction: NH4 NO3 –> N2 O + 2 H2 O?

• N2 O= mass of A
• NH4 NO3 = mass of B

Set up the formula so everything will cross out except molar mass B: 33g N2 O x (1 mole N2 O)/(44g N2 O) x (1 mole NH4 NO3 )/(1 mole N2 O) x (80g NH4 NO3 )/(1 mole NH4 NO3 ) = grams of NH4 NO3 .

60g = grams of NH4 NO3 !

## Lesson Summary

For any equation, you can calculate moles to moles, moles to mass, mass to moles and mass to mass by using simple equations and mole ratios. Use the following simple conversions:

• Moles A –> moles B –> mass B
• Mass A –> moles A –> moles B
• Mass A –> moles A –> moles B –> mass B

## Learning Outcome

You could be able to calculate moles to moles, moles to mass, mass to moles and mass to mass, using equations and ratios, after watching this lesson.

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