Question

A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molecular formulas.

Answer 1

Answer:

E.F= OH

M.F=$O_{2} H_{2}$

Explanation:

Empirical Formula

Step 1: Calculate mols of each element

0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H

*note leave extra sig figs for calculations

6.92gO(1 mol O/ 16 g O)=0.4325 mol O

Step 2: Identify which is the smallest mol

*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O

Step 3: Divide above calculations by the smallest mol

0.4325 mol O/0.4325= 1

0.4365 mol H/0.4325= 1.009 *rounds to 1

Step 4: use calculations as subscripts

Oxygen = 1 so the subscript will 1 (O)

Hydrogen = 1 so the subscript will 1 (H)

making E.F= OH

Molecular Formula:

Step 1: identify  molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

(oxygen mass from periodic table)+(hydrogen mass from periodic table)

16+1.008= 17.008

Step 2:Divide the molecular mass by the mass given by the emipirical formula.

$\frac{34.00}{17.008}= 1.999$ round to 2

Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula.  ANSWER: M.F=2(OH)- $O_{2} H_{2}$