18. Answer the following question that relate to the analysis of chemical compounds.

(a) A compound containing the elements C, H, N, and O is analyzed. When a 1.2359g sample is burned in excess oxygen, 2.241g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.048g of H.
(i) Determine the mass, in grams, of C in the 1.2359g sampleof the compound.
(ii) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84%. Determine the mass, in grams, of N, in the original 1.2359g sample of the compound.
(iii) Determine the mass, in grams, of O in the original 1.2359g sample of the compound.
(iv) Determine the empirical formula of the compound.

(I)

2.241 g CO2 x (1 mol CO2/molar mass CO2) x (atomic mass C/1 mol C) = g carbon.

(II)
1.2359 g sample x 0.2884 = g nitrogen.

(III)
1.2359 g sample - g C - g N -g H = grams O.

(IV)
Convert g C to percent C.
Convert g N to percent N.
Convert g H to percent H.
Convert g O to percent O.

Take a 100 g sample which will give you the grams of C, H, O, and N equal to the percents of each.
Convert g C, H, O, and N to mols by dividing by the atomic mass of C, H, O, and N respectively.
Convert to the ratio of small whole numbers. The easiest way to do this is to divide the smallest number of mols by itself (which gives it a value of 1.000). Divide all the other mole values by the same small number, then take the ratio of those whole numbers.
Post your work if you get stuck. Check my thinking. Check my work.

thanks so much for your help!

yes

Well, well, well, looks like you've got yourself a real chemistry party going on here! Let's break it down step by step and see if we can't find some answers.

(i) Alright, so we know that when the sample is burned, 2.241g of CO2 is formed. Now, CO2 is composed of one carbon atom and two oxygen atoms. So, to find the mass of carbon, we need to do a little math magic. Since the molar mass of CO2 is 44 g/mol, and there are 2.241g of CO2 formed, the mass of carbon in the sample is (2.241g CO2) x (1 mol CO2 / 44 g CO2) x (1 mol C / 1 mol CO2) x (12.01 g C / 1 mol C). Crunch those numbers and you should land right around... 0.129g of carbon!

(ii) Moving on to nitrogen, my dear element. We are told that the mass percent of nitrogen in the sample is 28.84%. Remember, mass percent is the mass of the element divided by the mass of the sample, multiplied by 100. So, we can set up an equation like this: (0.129g N / 1.2359g sample) x 100 = 28.84%. Cross multiply and divide, and you'll find yourself with... 0.356g of nitrogen!

(iii) Last, but not least, we have our good friend oxygen. To find the mass of oxygen, we simply subtract the masses of carbon, hydrogen, and nitrogen from the total mass of the sample. That's 1.2359g - 0.129g (from part (i)) - 0.048g (the mass of hydrogen) - 0.356g (from part (ii)). Crunch those numbers and you'll walk away with... 0.7029g of oxygen!

(iv) Now it's time to unlock the mystery of the empirical formula! To do this, we need to determine the ratios of the elements in the compound. From our previous calculations, we have 0.129g of carbon, 0.048g of hydrogen, 0.356g of nitrogen, and 0.7029g of oxygen.

Let's find the empirical formula step by step:
1. Convert the masses of each element to moles by dividing by their respective molar masses (C: 12.01 g/mol, H: 1.01 g/mol, N: 14.01 g/mol, O: 16.00 g/mol).
2. Divide each element's mole value by the smallest mole value.
3. Round the resulting values to the nearest whole number to get the subscripts in the empirical formula.

After following these steps, you should end up with your final answer, the empirical formula of the compound. Good luck, my chemist friend!

To determine the answers to these questions, we need to use the given information and perform calculations. Let's break down each question and go step by step.

(i) Determine the mass, in grams, of C in the 1.2359g sample of the compound:

1. We know that 2.241g of CO2 is formed when the sample is burned. From the balanced chemical equation of the combustion of carbon, we know that 1 mole of CO2 is formed for every 1 mole of C.

2. Calculate the number of moles of CO2 formed:
2.241g CO2 × (1 mol CO2 / 44.01g CO2) = 0.0509 mol CO2

3. Since there is 1 mol of C for every 1 mol of CO2, the number of moles of C is also 0.0509 mol.

4. Calculate the mass of C:
0.0509 mol C × (12.01g C / 1 mol C) = 0.611g C

Therefore, the mass of C in the 1.2359g sample is 0.611g.

(ii) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84%. Determine the mass, in grams, of N in the original 1.2359g sample of the compound:

1. The mass percent of N tells us that 28.84% of the compound's mass comes from N.

2. Calculate the mass of N:
(28.84 / 100) × 1.2359g = 0.3565g N

Therefore, the mass of N in the original 1.2359g sample is 0.3565g.

(iii) Determine the mass, in grams, of O in the original 1.2359g sample of the compound:

1. We can calculate the mass of O by subtracting the masses of C, H, and N from the total mass of the sample.

2. Calculate the mass of O:
Total mass - (mass of C + mass of H + mass of N)
1.2359g - (0.611g + 0.048g + 0.3565g) = 0.2204g O

Therefore, the mass of O in the original 1.2359g sample is 0.2204g.

(iv) Determine the empirical formula of the compound:

1. To find the empirical formula, we need the molar ratios of each element in the compound.

2. Calculate the moles of each element:
Moles of C = 0.0509 mol (as found in part (i))
Moles of H = 0.048g H / 1.01g/mol = 0.0475 mol H
Moles of N = 0.3565g N / 14.01g/mol = 0.0254 mol N
Moles of O = 0.2204g O / 16.00g/mol = 0.0138 mol O

3. Divide each mole value by the smallest mole value:
C: 0.0509 mol / 0.0138 mol ≈ 3.69
H: 0.0475 mol / 0.0138 mol ≈ 3.44
N: 0.0254 mol / 0.0138 mol ≈ 1.84
O: 0.0138 mol / 0.0138 mol = 1

4. Multiply the mole ratios by a whole number to get whole-number ratios:
C: 3.69 ≈ 4
H: 3.44 ≈ 3
N: 1.84 ≈ 2
O: 1

5. The empirical formula is obtained by writing the element symbols and their whole-number ratios:
C4H3N2O

Therefore, the empirical formula of the compound is C4H3N2O.