KEY2CHEM

Conservation of Matter in Chemical Reactions

During a chemical reaction or physical change, the law of conservation of mass must be obeyed. This means that the number of atoms of each type must be the same before and after the change. In a chemical formula, the subscripts represent the number of atoms of each type in a molecule. In a balanced chemical equation, the coefficients represent the relative numbers of particles that are consumed and created when the process occurs. The chemical equation relates what occurs on the atomic (microscopic) scale to what is observed on the larger (macroscopic) scale.


Example 1.

Using the smallest whole number coefficients, balance the equation below for the combustion of propane. What is the coefficient in front of \(\require{mhchem}\ce{H2O}\) in the balanced equation?

\(\require{mhchem}\ce{C3H8 + O2 -> CO2 + H2O}\)

A. \(2\)

B. \(4\)

C. \(8\)

 

 

Solution 

B. \(4\)

The number of atoms of each type must be balanced (the same) on each side of the equation. Using these coefficients, there are \(3 \text{ C}\) atoms on each side, \(8 \text{ H}\) atoms, and \(10 \text{ O}\) atoms, for an overall balanced equation of \(\require{mhchem}\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\). Therefore, all atoms have been conserved and the equation is balanced.

 


Example 2.

How many atoms of oxygen are present in \(5.99\text{ g}\) of \(\require{mhchem}\ce{Ba(NO3)2}\)?

A. \(8.28 \times10^{22} \text{ O atoms}\)

B. \(1.38 \times10^{22} \text{ O atoms}\)

C. \(4.23 \times10^{22} \text{ O atoms}\)

 

 

Solution

A. \(8.28 \times10^{22} \text{ O atoms}\)

Molar mass converts between mass and moles. The subscripts in the formula indicate the relative ratio of atoms. Avogadro's number converts between moles and atoms.

\(5.99\text{ g }\require{mhchem}\ce{Ba(NO3)2} \times \frac{1\text{ mol }\ce{Ba(NO3)2}}{261.35\text{ g }\ce{Ba(NO3)2}}\times \frac{6\text{ mol O}}{1\text{ mol }\ce{Ba(NO3)2}}\times \frac{6.022\times 10^{23}\text{ O atoms}}{1\text{ mol O atoms}} = 8.28 \times10^{22} \text{ O atoms}\)


Example 3.

How many moles of \(\require{mhchem}\ce{CH4}\) are needed to produce \(1.68\text{ g H}_2\text{O}\) by the following balanced chemical equation? 

\(\require{mhchem}\ce{CH4 + 2O2 -> CO2 + 2H2O}\)

A. \(4.66\times 10^{-2}\text{ mol}\)

B. \(9.33\times 10^{-2}\text{ mol}\)

C. \(8.40\times 10^{-1}\text{ mol}\)

 

 

Solution

A. \(4.66\times 10^{-2}\text{ mol}\)

The coefficients in the balanced equation can be used to convert between moles of reactants and products.   \(1.68\text{ g }\require{mhchem}\ce{H2O} \times \frac{1\text{ mol }\ce{H2O}}{18.016\text{ g }\ce{H2O}}\times \frac{1\text{ mol }\ce{CH4}}{2\text{ mol }\ce{H2O}}= 4.66 \times10^{-2} \text{ mol}\)