Compound A contains 5.2% nitrogen by mass, along with carbon, hydrogen, and oxygen. The combustion of 0.0850 g of compound A produced 0.224 g of CO2 and 0.0372 g of H2O. Calculate the empirical formula of compound A.

To calculate the empirical formula of compound A, we will follow these steps:

1. Determine the mass of carbon in the CO2 produced. 2. Determine the mass of hydrogen in the H2O produced. 3. Find the mass of oxygen in compound A by subtracting the mass of carbon, hydrogen, and nitrogen from the total mass of compound A. 4. Convert the mass of each element to moles by dividing by their respective atomic masses. 5. Find the simplest whole number ratio of moles of each element. 6. Write the empirical formula using these ratios.

Here are the detailed calculations:

Step 1: Determining the mass of carbon in CO2. The mass of carbon in CO2 can be found using the molar mass of CO2 (44.01 g/mol) and the fact that CO2 contains one carbon atom (molar mass of C is 12.01 g/mol). For every 44.01 g of CO2, there is 12.01 g of C. 0.224 g of CO2 * (12.01 g C / 44.01 g CO2) = 0.0611 g of C

Step 2: Determining the mass of hydrogen in H2O. The mass of hydrogen in H2O can be found using the molar mass of H2O (18.02 g/mol) and the fact that each H2O molecule contains two hydrogen atoms (molar mass of H is 1.01 g/mol). For every 18.02 g of H2O, there are 2.02 g of H. 0.0372 g of H2O * (2.02 g H / 18.02 g H2O) = 0.00415 g of H

Step 3: Calculating the mass of oxygen in compound A. We know that compound A contains carbon, hydrogen, and oxygen, along with 5.2% nitrogen. Let's find the mass of nitrogen first. The mass of nitrogen in the 0.0850 g of compound A is given by its percentage: 0.0850 g of compound A * 5.2% = 0.00442 g of N

Next, we subtract the mass of C, H, and N from the total mass of compound A to find the mass of oxygen: Total mass of compound A - (mass of C + mass of H + mass of N) = 0.0850 g - (0.0611 g + 0.00415 g + 0.00442 g) = 0.0850 g - 0.06967 g = 0.01533 g of O

Step 4: Converting the mass of each element to moles. Moles of C = 0.0611 g / 12.01 g/mol = 0.00509 mol C Moles of H = 0.00415 g / 1.01 g/mol = 0.00411 mol H Moles of N = 0.00442 g / 14.01 g/mol = 0.000316 mol N Moles of O = 0.01533 g / 16.00 g/mol = 0.000958 mol O

Step 5: Finding the simplest whole number ratio. To find the ratio, we divide each number of moles by the smallest number of moles among them: Ratio of C: \( \frac{0.00509}{0.000316} \approx 16.1 \) Ratio of H: \( \frac{0.00411}{0.000316} \approx 13.0 \) Ratio of N: \( \frac{0.000316}{0.000316} \approx 1 \) Ratio of O: \( \frac{0.000958}{0.000316} \approx 3.0 \)

These ratios suggest a whole number ratio of C16H13NO3. However, to confirm, we shouldn't expect a ratio as high as 16.1, which suggests the need to divide by a common divisor to get a whole number ratio that could potentially be more correct. We'll choose the smallest number of the significant digits in our moles, which seems to be two (from 0.000316 ~ 3.16), giving us approximately C16H13NO3 as the empirical formula.

Step 6: Writing the empirical formula. The empirical formula for compound A is approximately C16H13NO3.

Keep in mind that empirical formulas are typically the simplest whole number ratios, and these calculations are based on experimental data that might contain measurement errors. The ratios we got are close to whole numbers, but they are not exact, which suggests either some rounding or experimental error has occurred. A proper experiment should give ratios that are close to whole numbers without the need for significant adjustment.