Question

true or false:
a compound contains 38.7% c, 9.70% h, and 51.6% o. thus the empirical formula for the compound is ch₂o.

a true

b false

Explanation:

Step1: Assume 100g of compound

Masses: C = 38.7g, H = 9.70g, O = 51.6g.

Step2: Calculate moles

Moles of C: $\frac{38.7}{12.01} \approx 3.22$; Moles of H: $\frac{9.70}{1.008} \approx 9.62$; Moles of O: $\frac{51.6}{16.00} \approx 3.225$.

Step3: Divide by smallest mole (≈3.22)

C: $\frac{3.22}{3.22} = 1$; H: $\frac{9.62}{3.22} \approx 3$; O: $\frac{3.225}{3.22} \approx 1$. Wait, no—wait, original calculation: Wait, 38.7/12 ≈ 3.225, 9.7/1 ≈9.7, 51.6/16≈3.225. Then dividing by 3.225: C=1, H≈3, O=1? Wait no, wait the given empirical formula is CH₂O. Wait, my mistake: 38.7g C: moles = 38.7 / 12.01 ≈ 3.222; H: 9.70 / 1.008 ≈ 9.623; O: 51.6 / 16.00 ≈ 3.225. Now, divide each by the smallest (≈3.222): C: 1, H: 9.623 / 3.222 ≈ 3, O: 3.225 / 3.222 ≈ 1. Wait, that would be CH₃O? No, wait no—wait, no, the correct way: Wait, CH₂O has C:H:O = 1:2:1. Let's recalculate:

Wait, 38.7% C: moles C = 38.7 / 12 = 3.225; H: 9.7 / 1 = 9.7; O: 51.6 / 16 = 3.225. Now, divide by 3.225: C=1, H=9.7/3.225≈3, O=1. So the ratio is C:H:O ≈1:3:1, not 1:2:1. Wait, but wait, maybe I made a mistake. Wait, no—wait, the problem says "empirical formula is CH₂O". Let's check the ratios. For CH₂O, the mass percentages: C: (12)/(12+2+16) = 12/30 = 40%; H: 2/30 ≈6.67%; O: 16/30≈53.33%. But the given percentages are 38.7% C, 9.7% H, 51.6% O. Let's calculate the empirical formula from given percentages.

Moles: C: 38.7/12 ≈3.225; H:9.7/1=9.7; O:51.6/16=3.225. Now, divide by 3.225: C=1, H=9.7/3.225≈3, O=1. So the empirical formula is CH₃O? No, wait, 9.7 divided by 3.225 is approximately 3 (since 3.225*3=9.675, close to 9.7). So the ratio is C:H:O = 1:3:1, not 1:2:1. Therefore, the empirical formula is not CH₂O. Wait, but wait—wait, maybe I messed up the H calculation. Wait, 9.70% H: if we take 100g, H is 9.70g. Molar mass of H is 1.008 g/mol, so moles H = 9.70 / 1.008 ≈9.62 mol. Moles C: 38.7 / 12.01 ≈3.22 mol. Moles O: 51.6 / 16.00 ≈3.225 mol. Now, divide each by 3.22: C=1, H=9.62/3.22≈3, O=3.225/3.22≈1. So the empirical formula is CH₃O? No, that can't be. Wait, no—wait, the correct empirical formula calculation: let's do it properly.

Let’s take 100g sample:

• Mass of C = 38.7 g, molar mass C = 12.01 g/mol, moles C = 38.7 / 12.01 ≈ 3.222 mol

• Mass of H = 9.70 g, molar mass H = 1.008 g/mol, moles H = 9.70 / 1.008 ≈ 9.623 mol

• Mass of O = 51.6 g, molar mass O = 16.00 g/mol, moles O = 51.6 / 16.00 ≈ 3.225 mol

Now, divide each by the smallest number of moles (≈3.222):

• C: 3.222 / 3.222 = 1

• H: 9.623 / 3.222 ≈ 3 (since 3.222 * 3 = 9.666, close to 9.623)

• O: 3.225 / 3.222 ≈ 1

So the empirical formula is CH₃O? Wait, no, that's not right. Wait, the given empirical formula is CH₂O. So the ratios are C:1, H:2, O:1. But our calculation gives C:1, H:3, O:1. Therefore, the empirical formula is not CH₂O. Wait, but wait—maybe I made a mistake in the H moles. Wait, 9.70g H: 9.70 / 1.008 ≈9.62 mol. 3.222 mol C. 9.62 / 3.222 ≈3. So H is 3, not 2. Therefore, the empirical formula is CH₃O, not CH₂O. Therefore, the statement is False. Wait, but wait—wait, let's check the mass percentages for CH₂O. CH₂O: molar mass 30 g/mol. C: 12/30 = 40%, H: 2/30 ≈6.67%, O: 16/30≈53.33%. The given percentages are 38.7% C (less than 40%), 9.7% H (more than 6.67%), 51.6% O (less than 53.33%). So the H percentage is higher, which would mean more H moles, so the H ratio should be higher than 2. So the empirical formula should have more H than CH₂O. Therefore, the empirical formula is not CH₂O, so the statement is False.

Answer:

b. False