Question

note: round your ratio of moles to the nearest whole number as long as it is close to a whole number. in other words, you can round 1.992 up to 2, but you cant round 1.333 to 1. youll need to recognize common ratios, such as 1.333 being 4/3. for some compounds, the lowest number of atoms of an element might not be 1! if the lowest number of moles is four - thirds, you will need to multiply all ratios by 3 to get rid of the fraction. 2. a compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. the molecular weight of this compound is known to be approximately 140 g/mol. what is the empirical formula? (assume 100 grams of the substance is present.) what is the molecular formula? element mass (g) atomic mass mole mole ratio fractional subscript multiply by this number to get whole - number subscripts 0.20 0.25 0.33 0.50 0.66 0.75 empirical formula:

Explanation:

Step1: Calculate moles of each element

Assume 100 g of the compound. So, mass of C = 68.54 g, mass of H = 8.63 g, mass of O = 22.83 g.
Moles of C: $n_{C}=\frac{68.54\ g}{12.01\ g/mol}\approx 5.71\ mol$
Moles of H: $n_{H}=\frac{8.63\ g}{1.01\ g/mol}\approx 8.54\ mol$
Moles of O: $n_{O}=\frac{22.83\ g}{16.00\ g/mol}\approx 1.43\ mol$

Step2: Find the mole - ratio

Divide each number of moles by the smallest number of moles (1.43 mol in this case).
Ratio of C: $\frac{5.71}{1.43}\approx 4$
Ratio of H: $\frac{8.54}{1.43}\approx 6$
Ratio of O: $\frac{1.43}{1.43}=1$
The empirical formula is $C_{4}H_{6}O$.

Step3: Calculate the empirical - formula mass

Empirical - formula mass of $C_{4}H_{6}O=(4\times12.01)+(6\times1.01)+16.00 = 48.04 + 6.06+16.00=70.1\ g/mol$

Step4: Find the multiple (n)

Given the molecular weight is approximately 140 g/mol.
$n=\frac{\text{Molecular weight}}{\text{Empirical - formula mass}}=\frac{140\ g/mol}{70.1\ g/mol}\approx 2$

Step5: Determine the molecular formula

Multiply the subscripts in the empirical formula by n.
Molecular formula is $(C_{4}H_{6}O)\times2 = C_{8}H_{12}O_{2}$

Answer:

Empirical formula: $C_{4}H_{6}O$; Molecular formula: $C_{8}H_{12}O_{2}$