need help! thanks Review sheet A compound is found to be 65.45% C, 5.49% H, and
ID: 543144 • Letter: N
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need help! thanks
Review sheet A compound is found to be 65.45% C, 5.49% H, and 29.06% O by mass. In s separate experiment, the molar mass is found to be 110.111 g/mol. 1. Using the data above, find the empirical formula of the compound. a. Assuming 200.0 g of sample, calculate the number of moles of each element 1H Loo b. What is the ratio of C, H, and O atoms in the empirical formula unit? Determine the empirical formula c. 2. What is the molecular formula for the substance? a. What is the empirical formula mass? b. What is the molecular formula of the compound? 3. Review nomenclature of ALL ions that he gave in lecture. Version 1Explanation / Answer
1. Let us consider 100g of the compound is present and percentage to gram change
C = 65.45g , H = 5.49g , O = 29.06g
Now change the mass in mole. : Number of moles of each element (Molecular weight from periodic Table)
C = 65.45g/12.01g/mol = 5.45 mol
H = 5.49g/1.01g/mol = 5.43 mol
O = 29.06g/16g/mol = 1.81 mole
Divide by lowest no. : Ratio of C,H,O in empirical formula unit
C = 5.45 mol/1.81 mole = 3.01
H = 5.43 mol/1.81 mole = 3
O = 1.81 mole/1.81 mole = 1
Ratio of C,H,O = 3:3:1
Empirical formula : C3H3O
if assume 200 g of sample,
a. the no. of moles of each elements C,H,O will be twice 10.9, 10.86, 3.62 respectively.
b. Ratio of the each elements will be same: Ratio of C,H,O = 3:3:1 .
c. then the Empirical formula will be : C3H3O
2. The molecular formula is a multiple of the empirical formula. We were given the molecular weight of the molecule, 110.111 g/mol.
Number of empirical formula units in compound = 110.111 g/mol/56.06 g/mol = 2
Number of empirical formula units in compound = 2
so Molecular Formula of Compound = 2*C3H3O = C6H6O2
a. Empirical formula mass = 12.01*3 + 1.01*3 + 16*1 = 55.06
b. Molecular Formula of Compound = C6H6O2
3. Review nomenclature of All ions in determine molecular formula .
OH1-, OH1-, C6H42-
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