Fumaric acid, which occurs in many plants, contains, by mass, 41.4% carbon, 3.47% hydrogen, and 55.1%
oxygen. A 0.050-mol sample of this compound weighs 5.80 g. The molecular formula of this compound is can you show me the formula
I answered this below. It may not have been your post. If not, I'll post a link to the answer.
It WAS your post. Why are you posting the same question again. Didn't you look for the response?
http://www.jiskha.com/display.cgi?id=1254861796
To find the molecular formula of the compound, we need to determine the empirical formula first. The empirical formula gives the simplest ratio of atoms present in a compound.
1. Start by assuming a 100g sample of the compound. This assumption allows us to calculate the molar masses directly as percentages correspond to grams.
2. Convert the mass percentages to grams:
- Carbon: 41.4% = 41.4g
- Hydrogen: 3.47% = 3.47g
- Oxygen: 55.1% = 55.1g
3. Convert the masses to moles by dividing the mass by the molar mass:
- Carbon: 41.4g / 12.01 g/mol = 3.45 moles
- Hydrogen: 3.47g / 1.008 g/mol = 3.44 moles
- Oxygen: 55.1g / 16.00 g/mol = 3.44 moles
4. Divide the number of moles of each element by the smallest value to get the simplest mole ratio:
- Carbon: 3.45 moles / 3.44 moles = 1 mole
- Hydrogen: 3.44 moles / 3.44 moles = 1 mole
- Oxygen: 3.44 moles / 3.44 moles = 1 mole
Therefore, the empirical formula is C1H1O1. However, this formula does not exist. We need to determine the molecular formula by finding the ratio between the empirical formula mass and the experimental mass.
5. Calculate the empirical formula mass:
- C: 12.01 g/mol
- H: 1.008 g/mol
- O: 16.00 g/mol
Empirical formula mass = 12.01 g/mol + 1.008 g/mol + 16.00 g/mol = 29.02 g/mol
6. Divide the experimental mass by the empirical formula mass to find the multiplying factor:
- Experimental mass = 5.80 g
- Multiplying factor = 5.80 g / 29.02 g/mol = 0.200 mol
7. Multiply the subscripts in the empirical formula by the multiplying factor:
- C1H1O1 x 0.200 = C0.2H0.2O0.2
8. Multiply by 5 to remove decimal places if necessary:
- C0.2H0.2O0.2 x 5 = C1H1O1
Therefore, the molecular formula of the compound is C1H1O1, which is also the empirical formula.
Assume 100 g for the first part.
then you have 41.4 g of C, or xxx moles
then you have 3.47 g of H, or yyy moles H
and you have 55.1g of O, or zzz moles of O
Now, look at the xxxx,yyy, zzz numbers. Divide all of them by the LEAST number, you should be able to get a whole number ratio for the three elements.
That will give you the empirical formula.
Now, the mole mass is known:
molmass= 5.80/.050 g
divide the empirical formula mole mass into the known mole mass. That will give you a whole number ratio to multiply the subscripts in the empirical formula by. The result is the actual molecular formula.