Consider \(100.0-\mathrm{g}\) samples of two different compounds consisting only of carbon and oxygen. One compound contains \(27.2 \mathrm{~g}\) of carbon and the other has \(42.9 \mathrm{~g}\) of carbon. How can these data support the law of multiple proportions if \(42.9\) is not a multiple of \(27.2\) ? Show that these data support the law of multiple proportions.

First, we calculate the mass of oxygen in each compound using the given mass of the sample and the given mass of carbon. For compound 1: Mass of carbon: \(27.2 \mathrm{g}\) Mass of sample: \(100.0 \mathrm{g}\) Mass of oxygen = (Mass of sample) - (Mass of carbon) \(= 100.0 \mathrm{g} - 27.2 \mathrm{g}\) \(= 72.8 \mathrm{g}\) For compound 2: Mass of carbon: \(42.9 \mathrm{g}\) Mass of sample: \(100.0 \mathrm{g}\) Mass of oxygen = (Mass of sample) - (Mass of carbon) \(= 100.0 \mathrm{g} - 42.9 \mathrm{g}\) \(= 57.1 \mathrm{g}\)

Next, find the mass ratios between carbon (C) and oxygen (O) for each compound. For compound 1: Ratio: \(\frac{\text{mass of carbon}}{\text{mass of oxygen}}\) \(= \frac{27.2 \mathrm{g}}{72.8 \mathrm{g}}\) For compound 2: Ratio: \(\frac{\text{mass of carbon}}{\text{mass of oxygen}}\) \(= \frac{42.9 \mathrm{g}}{57.1 \mathrm{g}}\)

The data support the law of multiple proportions if the ratio between the mass ratios in both compounds is a simple whole-number multiple. \(\frac{\text{Ratio in compound 2}}{\text{Ratio in compound 1}}\) \(= \frac{\frac{42.9 \mathrm{g}}{57.1 \mathrm{g}}}{\frac{27.2 \mathrm{g}}{72.8 \mathrm{g}}}\) Now, cancel the units and simplify the fraction to see if it is a simple whole-number multiple. \(= \frac{\frac{42.9}{57.1}}{\frac{27.2}{72.8}}\) \(= \frac{42.9}{57.1} \times \frac{72.8}{27.2}\) \(= \frac{3083.32}{1547.32}\) The ratio is approximately equal to 2, indicating the data support the law of multiple proportions. Since the ratio between the mass ratios in both compounds is a simple whole-number multiple, the data provided supports the law of multiple proportions.