Questions: The empirical formula is a chemical formula that provides the smallest, whole-number ratio of elements in a compound. Using the mass of each element or the mass percentage of each element in a compound, the empirical formula can be determined. What is the empirical formula for a substance containing 28.6 grams of magnesium, Mg, 14.3 grams of carbon, C , and 57.1 grams of oxygen, O ?

Solution Steps

First, we need to convert the given masses of each element to moles using their respective molar masses.

• Molar mass of Mg: \(24.305 \, \text{g/mol}\)
• Molar mass of C: \(12.011 \, \text{g/mol}\)
• Molar mass of O: \(16.00 \, \text{g/mol}\)

\[ \text{Moles of Mg} = \frac{28.6 \, \text{g}}{24.305 \, \text{g/mol}} = 1.176 \, \text{mol} \]

\[ \text{Moles of C} = \frac{14.3 \, \text{g}}{12.011 \, \text{g/mol}} = 1.190 \, \text{mol} \]

\[ \text{Moles of O} = \frac{57.1 \, \text{g}}{16.00 \, \text{g/mol}} = 3.569 \, \text{mol} \]

Next, we find the smallest whole-number ratio by dividing each mole value by the smallest number of moles calculated.

\[ \text{Smallest number of moles} = 1.176 \, \text{mol} \]

\[ \frac{\text{Moles of Mg}}{1.176} = \frac{1.176}{1.176} = 1 \]

\[ \frac{\text{Moles of C}}{1.176} = \frac{1.190}{1.176} \approx 1.012 \]

\[ \frac{\text{Moles of O}}{1.176} = \frac{3.569}{1.176} \approx 3.035 \]

Since the ratios are very close to whole numbers, we can round them to the nearest whole number.

\[ \text{Ratio of Mg} = 1 \]

\[ \text{Ratio of C} \approx 1 \]

\[ \text{Ratio of O} \approx 3 \]

Final Answer