Calculate the empirical formula and molecular formula of a compound that contains only carbon, oxygen and hydrogen and has a molecular mass of approximately 90. g/mol. Upon combustion of 12.0g of the compund with an excess of oxygen, the yield was 1.2 g of water and 5.87 of carbon dioxide.

I tried to set the equation like this

12.0 --> 1.2 H2O + 5.87 CO2
but it said "with an excess of oxygen" and I was confuse on that...

I know how to find the molecular mass but in here it says 90. g/mol not just regular 90 grams, I don't know what to do with it. Do i have to change it to just grams or something? or just leave it.. also there was another problem similar to this. the molecular mass is "approximately 150 g/ml."

OK, you have 12.0 g sample.

On combustion, you obtained 1.20 g H2O and 5.87 g CO2, plus some oxygen (which may or may not have found its way into the H2O and CO2; however, that doesn't matter. We will find O another way).
1. Convert 1.20 g H2O to g H and convert 5.87 g CO2 to grams C. Now, grams oxygen = 12.0 - g H - g C = g oxygen. I think there is an error in the problem somewhere. These numbers don't look right to me.

2. Convert g C to moles. moles = g/molar mass.
Do the same for g H.
Do the same for g O.

3. Find the empirical formula by finding the ratio of these three elements to each other in small whole numbers. The easy way to do that is to divide the smallest number by itself, thereby assuring you a 1.000 for that element. Then divide the other numbers by the same small number. Round to whole numbers.

To calculate the empirical formula and molecular formula of the compound, you need to use the information given about the combustion reaction. Let's break down the steps to solve this problem.

Step 1: Calculate the amount of carbon and hydrogen in the compound.
From the given information, we have:
12.0 g of the compound → 5.87 g CO2
Using the molar masses, we can calculate the number of moles of CO2:
5.87 g CO2 × (1 mol CO2 / 44.01 g CO2) = 0.133 mol CO2

Since each mole of CO2 contains one mole of carbon, we have 0.133 mol of carbon.

Next, we need to calculate the amount of hydrogen. From the reaction equation, we can see that the combustion of the compound produces water (H2O). The given information tells us that the yield of water is 1.2 g.

1.2 g of water × (1 mol water / 18.02 g water) = 0.0666 mol water
Since each mole of water contains two moles of hydrogen, we have 0.133 mol of hydrogen.

Step 2: Calculate the amount of oxygen in the compound.
To find the amount of oxygen, we subtract the mass of carbon and hydrogen from the total mass of the compound.
Total mass of the compound = 12.0 g

Mass of carbon = 0.133 mol × (12.01 g/mol) = 1.60 g
Mass of hydrogen = 0.133 mol × (1.01 g/mol) = 0.134 g

Mass of oxygen = Total mass of the compound - Mass of carbon - Mass of hydrogen
Mass of oxygen = 12.0 g - 1.60 g - 0.134 g = 10.27 g

Step 3: Calculate the molar ratios.
Now that we have the amounts of each element, we need to find the simplest whole-number ratio between them. Divide the moles of each element by the smallest number of moles.

Carbon: 0.133 mol / 0.133 mol = 1 (rounded to the nearest whole number)
Hydrogen: 0.133 mol / 0.133 mol = 1 (rounded to the nearest whole number)
Oxygen: 10.27 g / 16.00 g/mol = 0.644 mol / 0.133 mol = 4.85 (rounded to the nearest whole number)

The empirical formula of the compound is CH4O5.

Step 4: Calculate the molecular formula.
To find the molecular formula, we need to determine the molar mass of the empirical formula. The given information states that the molecular mass is approximately 90 g/mol.

The empirical formula mass:
C: 12.01 g/mol × 1 = 12.01 g/mol
H: 1.01 g/mol × 4 = 4.04 g/mol
O: 16.00 g/mol × 5 = 80.00 g/mol

Adding these masses together, we get:
12.01 g/mol + 4.04 g/mol + 80.00 g/mol = 96.05 g/mol

To find the molecular formula, divide the molecular mass provided (90 g/mol) by the empirical formula mass (96.05 g/mol) to determine the multiplier:

90 g/mol ÷ 96.05 g/mol ≈ 0.936

Multiply the subscripts in the empirical formula by the multiplier to get the molecular formula:

C1H4O5 × 0.936 = C0.936H3.744O4.68

Rounding to the nearest whole number, we get the molecular formula as C1H4O5.

So, the empirical formula is CH4O5, and the molecular formula is C1H4O5.