January 23rd - Moles to Mass & Mass to Moles

Today, adding onto our last stochiometric lesson, we learned how to convert from grams to grams in a chemical equation. We must remember that converting from mass A to mass B must always go through moles A and moles B. Confused? This chart should clear it up:

Don’t worry, it’s easy! Converting from moles to mass only requires one additional step. In fact, it’s so simple that we can jump right into a few examples:

Ex. 1) Use the following formula to answer a) and b):

2H₂ + 0₂ à 2HOH

a) How many grams of HOH are produced when 2.50 moles of oxygen are used?

First, we simply convert from moles of oxygen to moles of HOH (we already know how to do this).

2.50 mol   x   (2/2)   =   5.00 mol of HOH

Next, we find the molar mass of oxygen. 2(1.0) + 16.0 = 18.0 g/mol. Then, like we learned in the last chapter, we convert!

5.00 mol   x   18.0 g/mol   =   90.0g of HOH

b) If 3.00 moles of HOH are produced, how many grams of oxygen must be consumed?

Using the same steps, we simply switch around the calculations.

3.00 mol   x   (1/2)   =   1.50 mol of O₂

We then find the molar mass of an oxygen molecule. 2(16.0) = 32.0 g/mol.

1.50 mol   x   32.0 g/mol   =   48.0g of O₂

Wow! That was fun! Now, here's something that'll get you in the mood for example 2:

Ex. 2) Iron (II) phosphate reacts with Tin (IV) nitride to produce Iron (II) nitride and Tin (IV) phosphate. How many moles of Tin (IV) nitride are needed to produce 0.500 mol of Iron (II) nitride? How many grams is this.

First, we must write the chemical formula.

2Fe₃(PO₄)₂ + Sn₃N₄ à 2Fe₃N₂ + Sn₃(PO₄)₄

Next, following what we did above, we convert from moles of Iron (II) nitride to Tin (IV) nitride.

0.500 mol   x   (1/2)   =   0.250 mol of Sn₃N₄

To find how many grams this is, we convert using the molar mass of Sn₃N₄. 3(118.7) + 4(14.0) = 412.1

0.250 mol   x   412.1 g/mol   =   103.0 grams of Sn₃N₄

Great job, everyone!

Here’s a sweet video for more practice:

Posted by Michael.