Question

An herbicide contains only CC, HH, ClCl, and NN. The complete combustion of a 250.0 mg250.0 mg sample of the herbicide in excess oxygen produced 261.5 mL261.5 mL of CO2CO2 and 152.5 mL152.5 mL of H2OH2O vapor at STP. A separate analysis determined the 250.0 mg250.0 mg sample contained 68.93 mg Cl68.93 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/mol257.16 g/mol. What is the molecular formula of the herbicide?

Answer 1

Mass of Sample = 250.0 mg

Mass of Cl in the sample = 68.93 mg

One mole of gas at STP contains 22.4 L of volume

Number of moles of CO2 = (261.5)/(22400) = 0.01167 moles of CO2

Number of moles of H2O = (152.5)/(22400) = 0.006808 moles of H2O

Mass of C in sample = 0.01167 * 12.01 = 140.15 mg

Mass of H in sample = 0.006808 * 2 * 1.008 = 13.72 mg

Mass of Nitrogen in sample = 250.0 - 68.93 - 140.15 - 13.72 = 27.2 mg

Moles of Cl in sample = 68.93 * 10^(-3)/35.45 = 0.0019444

Moles of nitrogen in sample = 27.2 * 10^(-3)/14 = 0.0019428

Hence the ratio will be C:H:Cl:N will be 0.01167:0.013616:0.001944:0.0019428, which will be 6:7:1:1

Hence the empirical formula will be C6H7NCl

Molar mass of empirical formula = 6 * 12.01 + 7 * 1.008 + 14 + 35.435 = 128.57

Molecular formula = 2 * empirical formula.

So, the molecular formula of herbicide will be C12H14N2Cl2

Note - Post any doubts/queries in comments section.

Answer 2

To determine the percent composition of the herbicide, we need to calculate the mass of each element present in the sample.

Mass of C: (261.5 mL CO2) x (1 mol CO2 / 22.4 L) x (12.011 g/mol C) = 147.37 mg C Mass of H: (152.5 mL H2O) x (1 mol H2O / 22.4 L) x (2 mol H / 1 mol H2O) x (1.008 g/mol H) = 136.43 mg H Mass of Cl: 68.93 mg Cl Mass of N: Total mass of sample - (mass of C + mass of H + mass of Cl) = 250.0 mg - (147.37 mg + 136.43 mg + 68.93 mg) = 106.27 mg N

Percent composition of C: (147.37 mg C / 250.0 mg) x 100% = 58.95% Percent composition of H: (136.43 mg H / 250.0 mg) x 100% = 54.57% Percent composition of Cl: (68.93 mg Cl / 250.0 mg) x 100% = 27.57% Percent composition of N: (106.27 mg N / 250.0 mg) x 100% = 42.51%

To determine the empirical formula of the herbicide, we need to convert the mass percentages to moles and find the simplest whole number ratio of the elements.

Assuming 100 g of the herbicide, the mass of each element in grams would be: C: 58.95 g H: 54.57 g Cl: 27.57 g N: 42.51 g

Converting these masses to moles: C: 58.95 g / 12.011 g/mol = 4.906 mol H: 54.57 g / 1.008 g/mol = 54.167 mol Cl: 27.57 g / 35.453 g/mol = 0.778 mol N: 42.51 g / 14.007 g/mol = 3.036 mol

The smallest whole number ratio of these moles is approximately 1C:11H:1Cl:3N. Therefore, the empirical formula of the herbicide is CH11ClN3.

To find the molecular formula, we need to know the molar mass of the empirical formula and the molar mass of the actual compound. The molar mass of CH11ClN3 is approximately 194.7 g/mol, which is calculated as follows:

(1 x 12.011 g/mol) + (11 x 1.008 g/mol) + (1 x 35.453 g/mol) + (3 x 14.007 g/mol) = 194.7 g/mol

The molecular formula of the herbicide can be determined by dividing the molar mass of the actual compound (257.16 g/mol) by the molar mass of the empirical formula (194.7 g/mol):

257.16 g/mol / 194.7 g/mol = 1.32

The molecular formula is approximately 1.32 times the empirical formula, which can be rounded up to CH14ClN4.