Question

An herbicide contains only C, H, Cl, and N.The complete combustion of a 150.0 mg sample of the herbicide in excess oxygen produced 156.9 mL of CO₂ and 91.52 mL of H₂O vapor at STP. A separate analysis determined the 150.0 mg sample contained 41.36 mg Cl. Determine the percent composition of the herbicide. Determine the empirical formula of the herbicide. The herbicide has a molar mass of 257.16 g/mol. What is the molecular formula of the herbicide?

Answer 1

The percent composition of the herbicide is approximately 66.4% carbon, 16.73% hydrogen, 27.57% chlorine, and 72.43% nitrogen.

Step-by-step explanation:

To determine the percent composition of the herbicide, we need to calculate the masses of carbon, hydrogen, chlorine, and nitrogen in a 150.0 mg sample. From the separate analysis, we know that the sample contains 41.36 mg of chlorine, which corresponds to 41.36/150.0 x 100% = 27.57% chlorine by mass. The empirical formula of the herbicide is C4H14ClN7. The molecular formula of the herbicide is approximately C8H28Cl2N14. The rest of the percent composition can be calculated as follows:

1. Mass of nitrogen = (150.0 - 41.36) mg = 108.64 mg
2. Mass of carbon = (108.64 x 61.0%) = 66.40 mg
3. Mass of hydrogen = (108.64 x 15.4%) = 16.73 mg

Therefore, the percent composition of the herbicide is approximately 66.4% carbon, 16.73% hydrogen, 27.57% chlorine, and 72.43% nitrogen.

To determine the empirical formula of the herbicide, we need to find the ratio of the elements in the compound. We can divide the mass or moles of each element by its atomic mass to find the ratio. Using the molar masses of carbon (12.01 g/mol), hydrogen (1.008 g/mol), chlorine (35.45 g/mol), and nitrogen (14.01 g/mol), we can calculate the moles of each element:

1. Moles of carbon = 66.40 mg / 12.01 g/mol = 5.53 x 10^-3 mol
2. Moles of hydrogen = 16.73 mg / 1.008 g/mol = 16.61 x 10^-3 mol
3. Moles of chlorine = 41.36 mg / 35.45 g/mol = 1.16 x 10^-3 mol
4. Moles of nitrogen = 108.64 mg / 14.01 g/mol = 7.75 x 10^-3 mol

Dividing the moles of each element by the smallest number of moles, which is 1.16 x 10^-3 mol, we get a composition ratio of approximately C4H14ClN7. Therefore, the empirical formula of the herbicide is C4H14ClN7.

To determine the molecular formula of the herbicide, we need to compare the empirical formula mass to the molar mass of the herbicide (257.16 g/mol). The empirical formula mass of C4H14ClN7 is (12.01 x 4) + (1.008 x 14) + (35.45 x 1) + (14.01 x 7) = 120.91 g/mol. Dividing the molar mass by the empirical formula mass, we get approximately 2.13. Therefore, the molecular formula of the herbicide is approximately 2 times the empirical formula, which is C8H28Cl2N14.