1.A compound consists of 65.45% C, 5.492% H, and 29.06% O on a mass basis and has a molecular mass of 110 g/mol. Determine the molecular formula of the compound.
2. What is the formula weight of iron (II) sulfate? Answer in units of g/mol.
3. What is the % nitrogen by weight in 2.368 mol of ammonium sulfide? Answer in units of %.
Take a 100 g sample. This gives you
65.45 g C
5.49 g H
29.06 g O
Convert to mols.
65.45/12 = ?
5.49/1 = ?
29.06/16 = ?
Now find the ratio of those elements to each other. The easy way to do that is to divide the smallest number by itself; then divide the other two numbers by the same small number.
2. FeSO4. All you need to do is to add atomic mass Fe to atomic mass of S to atomic mass of 4*oxygen.
3. Ammonium sulfide is (NH4)2S.
%N is same in the compound no matter how many moles you have.
%N = (2*atomic mass N/molar mass (NH4)2SO4)*100 = ?
1. To determine the molecular formula of the compound, we need to calculate the empirical formula first. The empirical formula represents the simplest whole-number ratio of the elements in a compound.
Step 1: Convert the mass percentages of each element into grams.
- Carbon (C): 65.45% of 110 g = 71.9955 g
- Hydrogen (H): 5.492% of 110 g = 6.05112 g
- Oxygen (O): 29.06% of 110 g = 31.966 g
Step 2: Convert the masses of each element into moles.
- Moles of C = 71.9955 g / atomic mass of C
- Moles of H = 6.05112 g / atomic mass of H
- Moles of O = 31.966 g / atomic mass of O
Step 3: Find the lowest whole-number ratio by dividing the moles of each element found above by the smallest number of moles.
- Divide moles of each element by the smallest number of moles obtained.
Step 4: Multiply the subscripts obtained in the previous step by a whole number, if the subscripts are not whole numbers, to obtain whole numbers (if necessary).
The resulting whole-number subscripts represent the empirical formula of the compound.
2. The formula weight (also called molar mass) of a compound is the sum of the atomic masses of its constituent elements, multiplied by the number of atoms for each element in the formula.
To calculate the formula weight of iron (II) sulfate (FeSO4), we need to know the atomic masses of iron (Fe), sulfur (S), and oxygen (O).
- Atomic mass of Fe = ...(Lookup in the periodic table)
- Atomic mass of S = ...(Lookup in the periodic table)
- Atomic mass of O = ...(Lookup in the periodic table)
Formula weight of FeSO4 = (Atomic mass of Fe) + (Atomic mass of S) + 4 * (Atomic mass of O)
3. To find the percentage of nitrogen by weight in ammonium sulfide, follow these steps:
Step 1: Determine the molar mass of ammonium sulfide (NH4)2S.
- Atomic mass of N = ...(Lookup in the periodic table)
- Atomic mass of H = ...(Lookup in the periodic table)
- Atomic mass of S = ...(Lookup in the periodic table)
Molar mass of (NH4)2S = 2 * (Atomic mass of N) + 8 * (Atomic mass of H) + (Atomic mass of S)
Step 2: Calculate the mass of nitrogen in 2.368 moles of ammonium sulfide.
- Mass of nitrogen = mole ratio * molar mass of nitrogen
- Mole ratio of nitrogen in (NH4)2S = 2 moles of N / 1 mole of (NH4)2S
Step 3: Calculate the percentage of nitrogen by weight.
- % nitrogen by weight = (Mass of nitrogen) / (Total mass of (NH4)2S) * 100%
To determine the molecular formula of the compound in question 1, we need to find the empirical formula first.
1. Find the moles of each element:
- Carbon (C): 65.45 g × (1 mol/12.01 g) = 5.454 mol
- Hydrogen (H): 5.492 g × (1 mol/1.01 g) = 5.441 mol
- Oxygen (O): 29.06 g × (1 mol/16.00 g) = 1.816 mol
2. Divide each mole value by the smallest mole value (1.816 mol):
- C: 5.454 mol ÷ 1.816 mol = 3
- H: 5.441 mol ÷ 1.816 mol ≈ 3
- O: 1.816 mol ÷ 1.816 mol = 1
3. The resulting ratio gives the empirical formula: C3H3O.
4. Determine the molecular formula by comparing the empirical formula mass to the given molecular mass (110 g/mol):
- Empirical formula mass: (3 × 12.01 g/mol) + (3 × 1.01 g/mol) + (1 × 16.00 g/mol) = 75.08 g/mol
- Molecular formula: (110 g/mol) ÷ (75.08 g/mol) = 1.464
Round the ratio to the nearest whole number, giving the molecular formula of C3H3O.
So, the molecular formula of the compound is C3H3O.
Moving on to question 2:
2. The formula for iron (II) sulfate is FeSO4. The atomic masses are:
- Iron (Fe): 55.85 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol (four oxygen atoms in the compound)
Calculating the formula weight:
(1 × 55.85 g/mol) + (1 × 32.07 g/mol) + (4 × 16.00 g/mol) = 151.90 g/mol
Therefore, the formula weight of iron (II) sulfate is 151.90 g/mol.
Finally, let's tackle question 3:
3. The molecular formula of ammonium sulfide is (NH4)2S.
To calculate the % nitrogen by weight in 2.368 mol of ammonium sulfide, we need to determine the molar mass of the compound and the molar mass of nitrogen.
The molar mass of ammonium sulfide is:
- (2 × 14.01 g/mol) + (8 × 1.01 g/mol) + (1 × 32.07 g/mol) = 68.15 g/mol
To find the molar mass of nitrogen:
- Nitrogen (N): 14.01 g/mol
Now, we can calculate the % nitrogen by weight:
- (2.368 mol × 14.01 g/mol) ÷ (68.15 g/mol) × 100% = 48.78%
Therefore, the % nitrogen by weight in 2.368 mol of ammonium sulfide is approximately 48.78%.