Percent Composition Answer Key (CHEM 1405)

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Percent Composition is the percent of each element, by mass, present in a chemical compound.  It is probably one of the most important pieces of information that can be determined about a chemical compound.  Percent composition data is usually gathered experimentally and it can be used for a variety of things including confirming the identity of a new chemical!  Percent composition can also be used to…  determine if a mine has a rich enough deposit of a mineral to make it worth the mining expense…  determine the amount of elements in foods and nutrients…  determine if two samples are of the same chemical compound, etc.  Please study the examples I have provided and then pursue some percent compositions on your own!

 

Example One:

From laboratory measurements, a sample of a pure compound is known to have a mass of 3.74 grams.  Analysis of the sample shows 1.10 g calcium, 0.880 g sulfur, and 1.76 g oxygen.  What is the percent composition of this compound?

$\frac{1.10gCa}{3.74g}\times100\%=$ 29.4% calcium

$\frac{0.880gSulfur}{3.74g}\times100\%=$ 23.4% sulfur

$\frac{1.76gOxygen}{3.74g}\times100\%=$ 47.1% oxygen

 

Example Two:

Calculate the percent composition of copper (II) nitrate, Cu(NO₃)₂, from its formula.

$1moleCu\times(\frac{63.55gCu}{1molCu})=63.55gCu$

$2moleN\times(\frac{14.01gN}{1molN})=28.02gN$

$6moleO\times(\frac{16.00gO}{1molO})=96.00gO$

Molar Mass:  $63.55g+28.02g+96.00g=187.57g/mol$

$\frac{63.55gCu}{187.57gCu(NO_3)_2}\times100\%=$ 33.9% Cu

$\frac{28.02gN}{187.57gCu(NO_3)_2}\times100\%=$ 14.9% N

$\frac{96.00gO}{187.57gCu(NO_3)_2}\times100\%=$ 51.2% O

 

You should note that the basic premise behind both examples of calculating percent composition is that:

 

 

Please show complete work.  Don’t forget to label your work with units and substances.

1.  A 9.14 gram sample contains 4.77 g carbon, 1.19 g hydrogen, and 3.18 g oxygen.  What is the percent composition, by mass, of this compound?

$\frac{4.77gC}{9.14g}\times100\%=$ 52.2% C

$\frac{1.19gH}{9.14g}\times100\%=$ 13.0% H

$\frac{3.18gO}{9.14g}\times100\%=$ 34.8% O

___52.2__ % carbon   

___13.0__ % hydrogen           

___34.8__ % oxygen

 

 

 

2.  A 2.85 gram sample contains 0.82 g magnesium, 1.62 g oxygen and the remainder is carbon.  What is the percent composition, by mass, of this compound?

2.85 g total - 0.82 g Mg - 1.62 g O = 0.41 g C

$\frac{0.41gC}{2.85g}\times100\%=$ 14% C

$\frac{0.82gMg}{2.85g}\times100\%=$ 29% Mg

$\frac{1.62gO}{2.85g}\times100\%=$ 56.8% O

___29___ % magnesium        

___56.8_ % oxygen  

___14___ % carbon

 

 

 

3.  Calculate the percent composition of aluminum sulfate from its formula, Al₂(SO₄)₃.

Al₂(SO₄)₃:

Al: 2 x 26.98 = 53.96
S:  3 x 32.07 = 96.21
O: 12 x 16.00 = 192.00
                           342.17 g/mol

$\frac{53.96gAl}{342.17gAl_2(SO_4)_3}\times100\%=$ 15.77% Al

$\frac{96.21gS}{342.17gAl_2(SO_4)_3}\times100\%=$ 28.12% S

$\frac{192.00gO}{342.17gAl_2(SO_4)_3}\times100\%=$ 56.112% O

__15.77_ % aluminum          

__28.12_ % sulfur               

_56.112_ % oxygen

 

 

 

4.  Calculate the percent composition of ammonium chloride from its formula.

NH₄Cl:
N: 1 x 14.01 = 14.01
H: 4 x 1.01   =   4.04
Cl: 1 x 35.45 = 35.45
                          53.50 g/mol

$\frac{14.01gN}{53.50gNH_4Cl}\times100\%=$ 26.19% N

$\frac{4.04gH}{53.50gNH_4Cl}\times100\%=$ 7.55% H

$\frac{35.45gCl}{53.50gNH_4Cl}\times100\%=$ 66.26% Cl

__26.19_ % nitrogen 

__7.55__ % hydrogen

__66.26_ % chlorine

 

 

 

Now, calculate percent composition and use it to answer the question…

5.  Two samples containing only iron and oxygen were analyzed.  Sample #1 weighed 56.0 grams of which 43.5 grams was iron and the rest oxygen.  Sample #2 weighed 177.0 grams of which 123.7 grams was iron and the rest oxygen.  Are the two samples the same compound?  Why/why not?

Sample #1	Sample #2
$\%Fe=\frac{43.5gFe}{56.0g}\times100\%=$ 77.7% Fe	$\%Fe=\frac{123.7gFe}{177.0g}\times100\%=$ 69.89% Fe

 

Sample #1 __77.7__ % iron                                                             

Sample #2 __69.89_ % iron

The two compounds are not the same.  If they were the same compound (having the same chemical formula), their percent compositions would have to be identical.

 

 

 

6.  A mining chemist has been sent a sample of copper-containing ore and has been told that due to mining costs, the sample must contain at least 30% copper to make the mine turn a profit.  He analyzes the 56.4 gram sample of ore and finds it contains 11.3 grams of sulfur, 22.6 grams of oxygen, and the rest is copper.  Based on his analysis, should the mining operation begin?  Why/why not?

56.4 g ore - 11.3 g S - 22.6 g O = 22.5 g Cu

$\frac{22.5gCu}{56.4g}\times100\%=$ 39.9% Cu

__39.9__ % copper

According to this analysis, the mining operation should proceed because the percent copper significantly exceeds the minimum 30% threshold for profitability.  

 

 

 

7.  Sodium phosphate is often an ingredient in detergents.  Calculate the percent composition of sodium phosphate and then determine how many grams of phosphorus are contained in a 908 gram sample of sodium phosphate.

Na₃PO₄:
Na: 3 x 22.99 = 68.97
P:    1 x 30.97 = 30.97
O:   4 x 16.00 = 64.00
                         163.94 g/mol

$\frac{68.97gNa}{163.94gNa_3PO_4}\times100\%=$ 42.07% Na

$\frac{30.97gP}{163.94gNa_3PO_4}\times100\%=$ 18.89% P

$\frac{64.00gO}{163.94gNa_3PO_4}\times100\%=$ 39.04% O

__42.07__ % sodium                        

__18.89__ % phosphorus                              

__39.04__ % oxygen

$908gNa_3PO_4\times\frac{18.89gP}{100gNa_3PO_4}=$ 172g P

 

__172__ grams of phosphorus