2. The Law of Definite Proportions:  A given chemical compound will always contain the same elements in the same proportions, by mass  This means that CO 2 (whether in solid or gas) always has the same % C and % O by mass

3.  How could we calculate this? Mass percent of C in CO 2 = M of C x 100% M of CO 2 Mass percent of C in CO 2 = 12.01g/mol C x100% 44.01 g/mol CO 2 Mass percent of C in CO 2 = 27.29 %

4.  Now what about O in CO 2 ? Mass percent of O in CO 2 = 2 x M of O x 100% M of CO 2 Mass percent of O in CO 2 = 2(16.00 g/mol) x 100% 44.01 g/mol Mass percent of O in CO 2 =72.71% Add them together – what do you get??

5.  Is it possible to find a different compound made of C and O where the % composition is different?  Yes! CO!!

6. Percentage Composition  The percentage composition of a compound refers to the relative mass of each element in the compound.  Percentage composition can be determined experimentally, and then be used to help identify the compound.

7. m X 100% m tot M X 100% M tot

8. Example: Determine the percent composition of sodium carbonate. Given: Na 2 CO 3 M Na = 2 X 22.99g/mol = 45.98 g/mol M C = 1 X 12.01 g/mol M O = 3 X 16.00 g/mol = 48.00 g/mol M Na 2 CO 3 = 45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 105.99 g/mol  The percent composition of Na 2 CO 3 is 43.38% Na, 11.33% C and 45.29% O % O = M O X 100% M Na2CO3 = 48.00 g/mol X 100% 105.99 g/mol = 45.29% % C = M C X 100% M Na2CO3 = 12.01 g/mol X 100% 105.99 g/mol = 11.33% % Na = M Na X 100% M Na2CO3 = 45.98 g/mol X 100% 105.99 g/mol = 43.38%

9.  Otherwise, it is possible to calculate the percent composition using raw mass data obtained in the lab

10. Example: The results of an experiment reveal that 2.5 g of hydrogen gas reacted completely to produce 22.5 g of water. What is the percent composition by mass of water? Given: m H = 2.5 g m H2O = 22.5 g m O = 22.5g - 2.5 g = 20.0 g % H = m H X 100% m H2O = 2.5 g X 100% 22.5 g = 11% % O = m O X 100% m H2O = 20.0 g X 100% 22.5 g = 88.9%  The percent composition of H 2 O is 11% H and 89 % O

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