Question

vitamin c has the formula cₓhᵧo_z. you burn 0.0400 g of the compound in a combustion analysis chamber and isolate 0.0600 g of co₂ and 0.0163 g of h₂o. what is the amount of c present in the original sample, in moles? 0.00136 mol co₂ (2 of 11)

Explanation:

Step1: Determine moles of \( CO_2 \)

Molar mass of \( CO_2 \) is \( 12.01 + 2\times16.00 = 44.01 \, g/mol \).
Moles of \( CO_2 = \frac{mass}{molar \, mass} = \frac{0.0600 \, g}{44.01 \, g/mol} \approx 0.00136 \, mol \) (wait, the given \( CO_2 \) mass might be 0.0600 g? Wait, the problem says "isolate 0.0600 g of \( CO_2 \)". Wait, but in the image, maybe typo? Wait, the key is: in combustion, each \( CO_2 \) has 1 C atom. So moles of C = moles of \( CO_2 \). Wait, no—wait, the sample is Vitamin C (\( C_xH_yO_z \)), burning it produces \( CO_2 \) and \( H_2O \). So moles of C in sample = moles of \( CO_2 \) produced.

Wait, let's recalculate: mass of \( CO_2 \) is 0.0600 g? Wait, the user's image: "isolate 0.0600 g of \( CO_2 \)". So moles of \( CO_2 = \frac{0.0600 \, g}{44.01 \, g/mol} \approx 0.00136 \, mol \). Since each \( CO_2 \) has 1 C, moles of C = moles of \( CO_2 \approx 0.00136 \, mol \)? Wait, but the problem says "What is the amount of C present in the original sample, in moles?"

Wait, maybe the \( CO_2 \) mass is 0.0600 g. Let's do it properly:

Molar mass of \( CO_2 \): \( M(CO_2) = 12.01 + 2 \times 16.00 = 44.01 \, g/mol \).

Moles of \( CO_2 = \frac{0.0600 \, g}{44.01 \, g/mol} \approx 0.00136 \, mol \).

Since each molecule of \( CO_2 \) contains one atom of C, the number of moles of C in the original sample is equal to the number of moles of \( CO_2 \) produced.

Step2: Confirm moles of C

Moles of C = moles of \( CO_2 \approx 0.00136 \, mol \). Wait, but the problem's given "0.00136 mol \( CO_2 \)"? Wait, maybe the \( CO_2 \) mass is 0.0600 g, so moles of \( CO_2 = 0.0600 / 44.01 ≈ 0.00136 \, mol \), so moles of C is also 0.00136 mol. Wait, but let's check the sample mass: 0.0400 g of the compound. But the key is the stoichiometry: C in sample → \( CO_2 \), so moles of C = moles of \( CO_2 \).

So the amount of C in moles is equal to the moles of \( CO_2 \), which is \( \frac{0.0600 \, g}{44.01 \, g/mol} ≈ 0.00136 \, mol \).

Answer:

\( \boxed{0.00136} \) (assuming the \( CO_2 \) mass is 0.0600 g, and moles of C = moles of \( CO_2 \))