Conservation of mass

Consider the balanced reaction from the previous lecture:

$$2{\rm C}_4{\rm H}_{10} + 13 {\rm O}_2 \longrightarrow 8{\rm CO}_2 + 10{\rm H}_2{\rm O}$$

In this example, suppose 2 mol of C$_4$H$_{10}$ are combined with 13 moles of O$_2$. What are the masses of the reactants and products?

Quantities in moles are converted to masses via the molar mass of each molecular species. The molar mass of a molecule is obtained by simply adding the molar masses of each of its atomic constituents. Thus, letting $M$ be the symbol for molar mass,

$$M_{{\rm C}_4{\rm H}_{10}} = 4M_{{\rm C}} + 10M_{{\rm H}} = 4\times 12.011 + 10\times 1.0079 = 58.123 {\rm\ g/mol}$$

$$M_{{\rm O}_2} = 2M_{{\rm O}} = 2.0\times 15.998 = 31.996 {\rm\ g/mol}$$

$$M_{{\rm CO}_2} = M_{{\rm C}} + 2M_{{\rm O}} = 12.011 + 2.0\times 15.998 = 44.007 {\rm\ g/mol}$$

$$M_{{\rm H}_2{\rm O}} = 2M_{{\rm H}} + M_{{\rm O}} = 2.0\times 1.0079 + 15.998 = 18.014 {\rm\ g/mol}$$

Then, 2 mol of C$_4$H$_{10}$ is equivalent to

$$2 {\rm\ mol\ } {\rm C}_4{\rm H}_{10} = (2 {\rm\ mol\ } {\rm ... ... 58.123 {\rm\ g/mol} = 116.25 {\rm\ g\ } {\rm C}_4{\rm H}_{10}$$

etc. Thus, the chemical equation can be written as a statement about masses of reactants and products:

$$116.25 {\rm\ g\ }{\rm C}_4{\rm H}_{10} + 415.95 {\rm\ g\ } {... ...52.06 {\rm\ g\ }{\rm CO}_2 + 180.14 {\rm\ g\ }{\rm H}_2{\rm O}$$

The total mass on the left side is 116.25 g + 415.95 g = 532.2 g, and the total mass on the right is 352.0 g + 180.14 g = 532.2 g. Thus, mass conservation is satisfied. The study of such mass relations in chemical reactions is called stoichiometry.