A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8635 g of CO2 and 0.1767 g of H2O. What is the empirical formula of the compound?

i got C3H3O2 IS IT RIGHT?

mass of C in sample = mass of C in CO2

mass of C in sample = (0.8635 g CO2 / 44 g/mol)(1 mol C / 1 mol CO2)(12 g/mol) = 0.2355 g

mass of H in sample = mass of H in H2O
mass of H in sample = (0.1767 g H2O / 18 g/mol)(2 mol H / 1 mol H2O)(1 g/mol) = 0.0196 g

mass of O in sample = total mass of sample - mass of C - mass of H
mass of O in sample = 0.4647 g - 0.2355 g - 0.0196 g = 0.2096 g
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Calculate for moles of C, H & O from the calculated masses:

moles of C = 0.2355 g / 12 g/mol = 0.0196 mol
moles of H = 0.0196 g / 1 g/mol = 0.0196 mol
moles of O = 0.2096 g / 16 g/mol = 0.0131 mol
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Divide this moles with the lowest value which is 0.0131 mol

C = 0.0196 / 0.0131 = 1.5
H = 0.0196 / 0.0131 = 1.5
O = 0.0131 / 0.0131 = 1
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Multiply by 2 to make the ratios a whole number:

C = 1.5 x 2 = 3
H = 1.5 x 2 = 3
O = 1 x 2 = 2

Empirical Formula is C3H3O2

Looks correct to me.

I didn't punch in your numbers, but the setups are correct, and if the CH ratio are 1.5 and O is 2, then you are correct.

Well, let's break this down! We can start by finding the number of moles of carbon, hydrogen, and oxygen in the given masses of CO2 and H2O.

The molar mass of CO2 (carbon dioxide) is 44.01 g/mol, so:

moles of carbon = mass of CO2 / molar mass of CO2
moles of carbon = 0.8635 g / 44.01 g/mol

Similarly, the molar mass of H2O (water) is 18.02 g/mol, so:

moles of hydrogen = mass of H2O / molar mass of H2O
moles of hydrogen = 0.1767 g / 18.02 g/mol

Now, let's find the number of moles of oxygen by subtracting the sum of the moles of carbon and hydrogen from the total moles of the compound:

moles of oxygen = total moles - (moles of carbon + moles of hydrogen)

Finally, we can use the moles of each element to get the empirical formula by dividing each value by the smallest one, to get a simple whole number ratio.

So, with all that math, the empirical formula of the compound is (drumroll, please)... not C3H3O2. Let's try again!

To determine the empirical formula of a compound, you need to find the ratio of the elements present in the compound.

First, you need to convert the mass of each element (carbon, hydrogen, and oxygen) to moles.

The molar mass of CO2 is calculated as follows:
CO2 = (1 × 12.01 g/mol) + (2 × 16.00 g/mol) = 44.01 g/mol.

The molar mass of H2O is calculated as follows:
H2O = (2 × 1.01 g/mol) + (1 × 16.00 g/mol) = 18.02 g/mol.

Next, calculate the moles of carbon, hydrogen, and oxygen present in the compound.

Moles of carbon = mass of CO2 / molar mass of CO2:
Moles of carbon = 0.8635 g / 44.01 g/mol.

Moles of hydrogen = mass of H2O / molar mass of H2O:
Moles of hydrogen = 0.1767 g / 18.02 g/mol.

Moles of oxygen = Total moles - (moles of carbon + moles of hydrogen):
Moles of oxygen = 0.4647 g / 31.99 g/mol - (moles of carbon + moles of hydrogen).

Finally, divide the number of moles for each element by the smallest number of moles calculated to find the mole ratio of the elements.

Let's calculate it:

For carbon:
Moles of carbon = 0.8635 g / 44.01 g/mol ≈ 0.0196 mol.

For hydrogen:
Moles of hydrogen = 0.1767 g / 18.02 g/mol ≈ 0.0098 mol.

For oxygen:
Moles of oxygen = [0.4647 g / 31.99 g/mol] - (0.0196 mol + 0.0098 mol).

Now, find the mole ratio:

Carbon:Hydrogen:Oxygen = 0.0196 mol : 0.0098 mol : (remaining moles of oxygen).

Determine the simplest whole-number ratio by dividing each value by the smallest value:

Carbon:Hydrogen:Oxygen ≈ 2:1:(remaining moles of oxygen).

In this case, the remaining moles of oxygen are negligible compared to carbon and hydrogen (likely due to rounding errors). So we ignore oxygen, and the empirical formula of the compound is C2H.