Answer:
Empirical formula - [tex]C_{7}H_{8}[/tex]
Explanation:
From the given,
Mass of [tex]CO_{2}[/tex] = 6.45 mg= 0.00645 g
Molar mass of carbon = 12 g/mol
Molar mass of [tex]CO_{2}[/tex] = 44 g/mol
Given mass of [tex]H_{2}O[/tex] = 1.51 mg= 0.00151 g
Molar mass of water = 18 g/mol
Molar mass of hydrogen = 1.0 g/mol
[tex]0.00645\times \frac{1molCO_{2}}{44gCO_{2}}\times \frac{1mol\,C}{1mol\,CO_{2}}= 1[/tex]
[tex]0.00151\times \frac{1molH_{2}O}{44gH_{2}O}\times \frac{2mol\,H}{1mol\,H_{2}O}= 1.4[/tex]
1:1.4
Those two values are multiplied by two.(molar ratio -7)
[tex]1:1.4\times 7= 7:8[/tex]
Therefore, empirical formula - [tex]C_{7}H_{8}[/tex]
