If a compound consists of 41.39 % carbon, 3.47 % hydrogen and 55.14 % oxygen what is the molecular formula if the molar mass is 116.07 g/mol?
c4h4o4
To determine the molecular formula of a compound, we need to find the empirical formula first, and then use the molar mass to find the molecular formula.
Step 1: Determine the empirical formula.
To find the empirical formula, we need to convert the percentages into moles. We can assume a 100g sample to make it easier to convert the percentages into grams.
For a 100g sample:
- Carbon (C) = 41.39 g
- Hydrogen (H) = 3.47 g
- Oxygen (O) = 55.14 g
Now, we need to convert the grams into moles using the molar masses of each element. The molar masses are:
- C: 12.01 g/mol
- H: 1.01 g/mol
- O: 16.00 g/mol
Converting grams to moles:
- Moles of C = 41.39 g / 12.01 g/mol = 3.45 mol
- Moles of H = 3.47 g / 1.01 g/mol = 3.44 mol
- Moles of O = 55.14 g / 16.00 g/mol = 3.45 mol
To find the simplest whole number ratio of the elements, we divide each mole value by the smallest mole value, which in this case is 3.44 mol.
Dividing by 3.44:
- C: 3.45 mol / 3.44 mol = 1.007 mol ≈ 1
- H: 3.44 mol / 3.44 mol = 1 mol ≈ 1
- O: 3.45 mol / 3.44 mol = 1.003 mol ≈ 1
The empirical formula is CH3O.
Step 2: Determine the molecular formula.
To find the molecular formula, we need to know the molar mass of the compound. In this case, the molar mass is given as 116.07 g/mol.
The molar mass of the empirical formula CH3O can be calculated as:
- Carbon (C): 12.01 g/mol × 1 = 12.01 g/mol
- Hydrogen (H): 1.01 g/mol × 3 = 3.03 g/mol
- Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol
Adding up the molar masses:
12.01 g/mol + 3.03 g/mol + 16.00 g/mol = 31.04 g/mol
Now, we can calculate the empirical formula's molar mass (31.04 g/mol) multiple in order to obtain a value close to the given molar mass (116.07 g/mol).
116.07 g/mol / 31.04 g/mol = 3.73
Therefore, the molecular formula is approximately (CH3O)3 or C3H9O3.
To determine the molecular formula of a compound, we need to calculate the empirical formula first. The empirical formula represents the simplest whole-number ratio of atoms present in a compound.
Step 1: Convert the percentages to grams.
For a 100g sample of the compound:
- Carbon: 41.39g
- Hydrogen: 3.47g
- Oxygen: 55.14g
Step 2: Convert the grams to moles.
To do this, divide each element's mass by its atomic molar mass.
- Carbon: 41.39g / 12.01 g/mol = 3.448 mol of C
- Hydrogen: 3.47g / 1.01 g/mol = 3.436 mol of H
- Oxygen: 55.14g / 16.00 g/mol = 3.446 mol of O
Step 3: Find the simplest whole-number ratio.
To achieve this, divide each number of moles by the smallest number of moles (which, in this case, is 3.436 mol of H).
- Carbon: 3.448 mol / 3.436 mol ≈ 1
- Hydrogen: 3.436 mol / 3.436 mol = 1
- Oxygen: 3.446 mol / 3.436 mol ≈ 1
Therefore, the empirical formula is CH2O.
Step 4: Calculate the empirical formula mass.
The empirical formula mass can be obtained by summing the atomic masses of the elements in the empirical formula.
- Carbon: 1 × 12.01 g/mol = 12.01 g/mol
- Hydrogen: 2 × 1.01 g/mol = 2.02 g/mol
- Oxygen: 1 × 16.00 g/mol = 16.00 g/mol
The empirical formula mass is (12.01 g/mol) + (2.02 g/mol) + (16.00 g/mol) = 30.03 g/mol.
Step 5: Determine the multiple of the empirical formula mass needed to match the given molar mass.
Divide the molar mass of the compound (116.07 g/mol) by the empirical formula mass (30.03 g/mol).
116.07 g/mol ÷ 30.03 g/mol ≈ 3.87
Hence, the molecular formula is approximately 3.87 times the empirical formula, which gives us: C3.87H7.75O3.87.
For simplicity, we can multiply all subscripts by 4 to get a whole number ratio, resulting in the molecular formula C15H31O15.
Take a 100 g sample. That will give you
3.47 g H
41.39 g C
55.14 g O.
Convert grams to moles.
41.39/12.01 = ?? moles C.
3.47/1 = ?? moles H.
55.14/16 = ?? moles O.
Now you want to find the ratio of the C, H, and O to each other in small whole numbers. The easy way to do that is to divide the smallest value by itself (thereby assuring 1.000 for that value), then divide the other values by the same small value, round to whole numbers, and that gives you the empirical formula. Convert to molecular formula the same way as the previous problem I worked for you.