3. Find the empirical formula for a compound that consists of aluminum and chlorine in which the aluminum is 20.2 % by mass.
4. A sample contains 71.65% CL, 24.27% C nad 4.07% H. the molecular weight is known to be 98.96 g/mol. What are the empirical and molecular formulas?
3)
Take a 100 g sample. This gives you
20.2 g Al
79.8 g O
Convert to moles.
20.2/atomic mass Al = ?
79.8/atomic mass O = ?
Now find the ratio of the two elements to each other with the smallest number being 1.00. Don't round to whole numbers too much.
To find the empirical formula for a compound, you need to determine the ratio of elements present in the compound. This can be done by finding the percent composition of each element.
3. Finding the empirical formula for a compound of aluminum and chlorine:
1. Assume we have a 100 g sample of the compound.
2. Given that aluminum is 20.2% by mass, we have 20.2 g of aluminum.
3. Given that chlorine makes up the remaining mass, we have 100 g - 20.2 g = 79.8 g of chlorine.
4. Convert the mass of each element to moles using their respective molar masses.
- Aluminum (Al) has a molar mass of 26.98 g/mol, so we have 20.2 g / 26.98 g/mol = 0.749 mol Al.
- Chlorine (Cl) has a molar mass of 35.45 g/mol, so we have 79.8 g / 35.45 g/mol = 2.25 mol Cl.
5. Divide the number of moles of each element by the smallest number of moles to obtain a simplified ratio.
- Dividing by 0.749 (the number of moles of Al), we get 0.749 mol Al / 0.749 mol Cl = 1 mol Al / 3 mol Cl (approximately).
6. The empirical formula for the compound is AlCl₃.
4. To find both the empirical and molecular formulas of a compound:
1. Assume we have a 100 g sample of the compound, so we have 71.65 g Cl, 24.27 g C, and 4.07 g H.
2. Convert the mass of each element to moles using their respective molar masses.
- Chlorine (Cl) has a molar mass of 35.45 g/mol, so we have 71.65 g / 35.45 g/mol = 2.02 mol Cl.
- Carbon (C) has a molar mass of 12.01 g/mol, so we have 24.27 g / 12.01 g/mol = 2.02 mol C.
- Hydrogen (H) has a molar mass of 1.01 g/mol, so we have 4.07 g / 1.01 g/mol = 4.03 mol H.
3. Divide the number of moles of each element by the smallest number of moles to obtain a simplified ratio.
- Dividing by 2.02 (the number of moles of Cl and C), we get 2.02 mol Cl / 2.02 mol Cl = 1 mol Cl, and 2.02 mol C / 2.02 mol Cl = 1 mol C.
- Dividing by 2.02 (the number of moles of Cl and H), we get 2.02 mol Cl / 2.02 mol H = 1 mol Cl, and 4.03 mol H / 2.02 mol H = 2 mol H.
4. The empirical formula for the compound is CHCl.
5. To determine the molecular formula, we need to know the molar mass of the compound.
- Given that the molecular weight is 98.96 g/mol, we divide it by the empirical formula weight to find the molecular formula weight.
- The empirical formula weight of CHCl is 1(12.01) + 1(1.01) + 1(35.45) = 48.47 g/mol.
- Dividing the molecular weight (98.96 g/mol) by the empirical formula weight (48.47 g/mol), we get approximately 2.04.
6. Multiply the subscripts in the empirical formula by 2 (the number obtained from the division) to get the molecular formula.
- The molecular formula for CHCl is 2CHCl, which can be simplified to C₂H₂Cl₂.