## Q. 4.1

Writing and Balancing an Equation: The Combustion of a Carbon–Hydrogen–Oxygen Compound

Liquid triethylene glycol is used as a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for the combustion of this compound in a plentiful supply of oxygen. A ball-and-stick model of triethylene glycol is shown here.

## Verified Solution

Analyze

We deduce the formula of triethylene glycol from the molecular model. (Refer to the color scheme given on the inside back cover.) We see 6 C atoms (black), 4 O atoms (red), and 14 H atoms. The formula is $C_{6}H_{14}O_{4}$. When a carbon–hydrogen–oxygen compound is burned in excess oxygen, $O_{2}$, the products are $CO_{2}$ and $H_{2}O$.

Solve

Having identified the reactants and products, we write down an unbalanced chemical expression for the reaction, showing all reactants and products, and then we balance the expression with respect to each kind of atom

Starting expression:          $C_{6}H_{14}O_{4} + O_{2} → CO_{2} + H_{2}O$

Balance C:        $C_{6}H_{14}O_{4} + O_{2} → 6 CO_{2} + H_{2}O$

Balance H:        $C_{6}H_{14}O_{4} + O_{2} → 6 CO_{2} + 7 H_{2}O$

At this point, the right side of the expression has 19 O atoms (12 in six $CO_{2}$ molecules and 7 in seven $H_{2}O$ molecules), and the left side, only 4 O atoms (in $C_{6}H_{14}O_{4}$). To obtain 15 more O atoms requires a fractional coefficient of 15/2 for $O_{2}$.

Balance O:          $C_{6}H_{14}O_{4} + \frac{15}{2}O_{2} → 6 CO_{2} + 7 H_{2}O$     (balanced)

To remove the fractional coefficient, multiply all coefficients by two:

2 $C_{6}H_{14}O_{4} + 15 O_{2} → 12 CO_{2} + 14 H_{2}O$                          (balanced)

Assess

To check that the equation is balanced, determine the numbers of C, H, and O atoms that appear on the each side of the equation.

Left:      (2 × 6) = 12 C;    (2 × 14) = 28 H;     [(2 × 4) + (15 × 2)] = 38 O

Right:    (12 × 1) = 12 C;    (14 × 2) = 28 H;    [(12 × 2) + (14 × 1)] = 38 O