A compound is 54.52% C, 9.15% H, and 36.32% O by mass. What is the empirical formula?
The molecular mass of the compound is 132 amu. What is the molecular formula?
Take a 100 g sample, convert percent to grams.
Convert grams each to moles.
Then find the ratio as in your other problems.
Determine the mass of the empirical formula, then find (molar mass/empirical mass) = n and round to a whole numbers.
(empirical formula)n is the molecular formula. Post your work if you get stuck.
To determine the empirical formula, we need to find the simplest whole number ratio of the elements present in the compound.
Step 1: Convert the percentages to grams.
For a 100g sample of the compound, we have:
C = 54.52g, H = 9.15g, O = 36.32g.
Step 2: Convert the grams to moles.
Use the molar masses of each element to convert grams to moles:
C: 54.52g C x (1 mol C/12.01g C) = 4.54 mol C
H: 9.15g H x (1 mol H/1.01g H) = 9.06 mol H
O: 36.32g O x (1 mol O/16.00g O) = 2.27 mol O
Step 3: Find the smallest whole number ratio.
Divide each element's moles by the smallest number of moles calculated (2.27 mol O):
C: 4.54 mol C / 2.27 mol O = 2
H: 9.06 mol H / 2.27 mol O = 4
O: 2.27 mol O / 2.27 mol O = 1
The empirical formula is CH2O, indicating that the ratio of carbon, hydrogen, and oxygen atoms is 1:2:1.
To find the molecular formula, we need to know the molecular mass of the compound. Given that the molecular mass is 132 amu, we can calculate the molecular formula by comparing it to the empirical formula.
Step 4: Calculate the empirical formula mass.
Sum the atomic masses of carbon, hydrogen, and oxygen in the empirical formula:
CH2O: (1 x 12.01) + (2 x 1.01) + (1 x 16.00) = 30.03 amu
Step 5: Determine the ratio between the molecular mass and empirical formula mass.
Divide the molecular mass (132 amu) by the empirical formula mass (30.03 amu):
132 amu / 30.03 amu = 4.39
Step 6: Multiply the subscripts in the empirical formula by the ratio obtained in Step 5.
Multiply each subscript in the empirical formula by 4.39:
C: 1 x 4.39 = 4.39
H: 2 x 4.39 = 8.78
O: 1 x 4.39 = 4.39
Rounding each value to the nearest whole number, we get the molecular formula: C4H9O4.
Therefore, the empirical formula of the compound is CH2O, and its molecular formula is C4H9O4.