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