A sample has a molar mass of 162g/mol. When a sample of mass 0.385g was burned , 1.072g of CO2 ,0.307g H2O and 0.068g of nitrogen were produced. What is empirial and molecular formular of a sample . Write equation for combustion
To determine the empirical formula of the sample, we need to find the moles of each element present in the compound. We can do this by dividing the mass of each element by its molar mass.
1. CO2:
- The molar mass of CO2 is 44.01 g/mol.
- The mass of CO2 produced is 1.072 g.
- Moles of CO2 = (mass of CO2) / (molar mass of CO2) = 1.072 g / 44.01 g/mol = 0.0244 mol.
2. H2O:
- The molar mass of H2O is 18.02 g/mol.
- The mass of H2O produced is 0.307 g.
- Moles of H2O = (mass of H2O) / (molar mass of H2O) = 0.307 g / 18.02 g/mol = 0.017 mol.
3. Nitrogen (N):
- The molar mass of N is approximately 14.01 g/mol.
- The mass of N produced is 0.068 g.
- Moles of N = (mass of N) / (molar mass of N) = 0.068 g / 14.01 g/mol = 0.0048 mol.
Now, we need to find the mole ratio of each element. Divide the moles of each element by the smallest number of moles (in this case, N) to get the mole ratios.
Moles of CO2: 0.0244 mol / 0.0048 mol = 5.083
Moles of H2O: 0.017 mol / 0.0048 mol = 3.542
Moles of N: 0.0048 mol / 0.0048 mol = 1.0
The mole ratios are approximately CO2: 5, H2O: 4, N: 1.
To simplify the ratios, we can divide all the values by the smallest mole ratio (N: 1), which gives us a ratio of approximately CO2: 5, H2O: 4, N: 1.
Therefore, the empirical formula of the compound is C5H4N.
To determine the molecular formula of the compound, we need the molar mass of the compound. The molar mass is given as 162 g/mol.
To calculate the molecular formula, we divide the molar mass of the compound by the molar mass of the empirical formula:
Molar mass of empirical formula = (5 × molar mass of C) + (4 × molar mass of H) + (1 × molar mass of N)
= (5 × 12.01 g/mol) + (4 × 1.01 g/mol) + (1 × 14.01 g/mol)
= 60.05 g/mol + 4.04 g/mol + 14.01 g/mol
= 78.1 g/mol
Now, divide the molar mass of the compound (162 g/mol) by the molar mass of the empirical formula (78.1 g/mol) to get the molecular formula ratio:
Molecular formula ratio = (molar mass of compound) / (molar mass of empirical formula) = 162 g/mol / 78.1 g/mol = 2.07.
The molecular formula ratio is approximately 2.07. Since we cannot have fractional subscripts in a formula, we need to round this ratio to the nearest whole number.
Rounded to the nearest whole number, the molecular formula is approximately 2 times the empirical formula. Therefore, the molecular formula is C10H8N.
Finally, to write the equation for combustion, we combine the empirical formula with oxygen (O2) to show the reactants and products involved in the combustion process.
C10H8N + O2 → CO2 + H2O + N2
To determine the empirical and molecular formula of a sample, we need to follow a few steps:
Step 1: Calculate the moles of each element present in the sample.
To calculate the moles of each element, we divide the mass of each element by its molar mass.
Given:
Mass of CO2 produced = 1.072g
Molar mass of CO2 = 44g/mol
Moles of carbon (C) in CO2 = (1.072g CO2 / 44g/mol CO2) = 0.0244 mol C
Moles of oxygen (O) in CO2 = (2 * 1.072g CO2 / 44g/mol CO2) = 0.0488 mol O
Mass of H2O produced = 0.307g
Molar mass of H2O = 18g/mol
Moles of hydrogen (H) in H2O = (2 * 0.307g H2O / 18g/mol H2O) = 0.0341 mol H
Moles of oxygen (O) in H2O = (0.307g H2O / 18g/mol H2O) = 0.0171 mol O
Mass of nitrogen produced = 0.068g
Molar mass of nitrogen (N2) = 28g/mol
Moles of nitrogen (N) in N2 = (0.068g N2 / 28g/mol N2) = 0.0024 mol N
Step 2: Find the empirical formula.
The empirical formula shows the ratio of atoms in the simplest whole-number form. To determine it, we divide the number of moles of each element by the smallest number of moles.
Dividing the moles by the smallest number of moles (0.0024 mol N), we get:
Carbon (C) = 0.0244 mol C / 0.0024 mol N = 10
Hydrogen (H) = 0.0341 mol H / 0.0024 mol N = 14
Oxygen (O) = 0.0488 mol O / 0.0024 mol N = 20
Therefore, the empirical formula is C10H14O20.
Step 3: Find 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 162g/mol.
The molar mass of the empirical formula (C10H14O20) can be calculated as follows:
(10 * molar mass of carbon) + (14 * molar mass of hydrogen) + (20 * molar mass of oxygen)
Substituting the respective molar masses:
(10 * 12) + (14 * 1) + (20 * 16) = 120 + 14 + 320 = 454
To find the molecular formula, we divide the molar mass of the compound (162g/mol) by the molar mass of the empirical formula (454g/mol). This will give us the multiple of the empirical formula in the molecular formula.
Molecular formula = (162g/mol) / (454g/mol) = 0.357
Multiplying each subscripts in the empirical formula by 0.357:
C (10 * 0.357) = 3.57 ≈ 4
H (14 * 0.357) = 5.01 ≈ 5
O (20 * 0.357) = 7.14 ≈ 7
Therefore, the molecular formula of the sample is C4H5O7.
Step 4: Write the combustion equation.
The combustion equation shows the reaction between the sample and oxygen to produce carbon dioxide, water, and nitrogen.
C4H5O7 + O2 -> 4CO2 + 2H2O + N2
Please note that this is a balanced equation, as the coefficients in front of each molecule represent the number of molecules involved in the reaction.