Questions: Question 2 1 pts A compound comprised of calcium and nitrogen weighs 29.6 g and contains 24.0 g of calcium. What is the empirical formula? CaN3 CaN Ca3 N2 Ca2 N2 Question 3 1 pts A 170.00 g sample of an unidentified compound contains 29.84 g sodium, 67.49 g chromium, and 72.67 oxygen. What is the compound's empirical formula? NaCrO3.5 Na2 Cr2 O7 Na2 CrO7 NaCrO7

Solution Steps

For the first question, we have a compound containing calcium (Ca) and nitrogen (N). The compound weighs 29.6 g, with 24.0 g of calcium. First, we need to find the moles of each element.

• Moles of calcium: \[ \text{Moles of Ca} = \frac{24.0 \, \text{g}}{40.08 \, \text{g/mol}} = 0.5992 \, \text{mol} \]

• Moles of nitrogen: \[ \text{Mass of N} = 29.6 \, \text{g} - 24.0 \, \text{g} = 5.6 \, \text{g} \] \[ \text{Moles of N} = \frac{5.6 \, \text{g}}{14.01 \, \text{g/mol}} = 0.3997 \, \text{mol} \]

Next, we find the simplest whole number ratio of moles of calcium to moles of nitrogen.

• Ratio of Ca to N: \[ \frac{0.5992}{0.3997} \approx 1.5 \] \[ \frac{0.3997}{0.3997} = 1 \]

To convert to whole numbers, multiply both by 2:

• Ca: \(1.5 \times 2 = 3\)
• N: \(1 \times 2 = 2\)

The empirical formula is based on the whole number ratio of moles: \[ \text{Empirical formula: } \mathrm{Ca}_3\mathrm{N}_2 \]

For the second question, we have a compound containing sodium (Na), chromium (Cr), and oxygen (O). The sample weighs 170.00 g, with the following masses:

• Moles of sodium: \[ \text{Moles of Na} = \frac{29.84 \, \text{g}}{22.99 \, \text{g/mol}} = 1.2975 \, \text{mol} \]

• Moles of chromium: \[ \text{Moles of Cr} = \frac{67.49 \, \text{g}}{51.996 \, \text{g/mol}} = 1.2970 \, \text{mol} \]

• Moles of oxygen: \[ \text{Moles of O} = \frac{72.67 \, \text{g}}{16.00 \, \text{g/mol}} = 4.5419 \, \text{mol} \]

Find the simplest whole number ratio of moles of each element.

• Ratio of Na: \[ \frac{1.2975}{1.2970} \approx 1 \]

• Ratio of Cr: \[ \frac{1.2970}{1.2970} = 1 \]

• Ratio of O: \[ \frac{4.5419}{1.2970} \approx 3.5 \]

To convert to whole numbers, multiply all by 2:

• Na: \(1 \times 2 = 2\)
• Cr: \(1 \times 2 = 2\)
• O: \(3.5 \times 2 = 7\)

The empirical formula is based on the whole number ratio of moles: \[ \text{Empirical formula: } \mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7 \]

Final Answer