A sample of a compound containing only carbon and hydrogen is burned, producing 157 mg CO2 and 42.8 mg H2O. What is the empirical formula?

Question

A sample of a compound containing only carbon and hydrogen is burned, producing 157 mg CO2 and 42.8 mg H2O. What is the empirical formula?

1 Answer

Impact of this question

13148 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-06T00:35:06

Combustion analysis!

$157 m g \cdot \frac{10^{- 3} g}{m g} \cdot \frac{C O_{2}}{44.01 g} \approx 3.57 \cdot 10^{- 3} m o l$ $157mg * (10^-3g)/(mg) * (CO_2)/(44.01g) approx 3.57*10^-3mol$

$42.8 m g \cdot \frac{10^{- 3} g}{m g} \cdot \frac{H_{2} O}{18.02 g} \cdot \frac{2 H}{H_{2} O} \approx 4.75 \cdot 10^{- 3} m o l$ $42.8mg * (10^-3g)/(mg) * (H_2O)/(18.02g)* (2H)/(H_2O) approx 4.75*10^-3mol$

If you divide the largest molar mass by the smallest...

$C_{1.33} H$ $C_1.33H$

Annoying... multiply by three so integers are the subscripts:

$C_{4} H_{3}$ $C_4H_3$

Science

Math

Humanities

... and beyond

A sample of a compound containing only carbon and hydrogen is burned, producing 157 mg CO2 and 42.8 mg H2O. What is the empirical formula?

1 Answer

Related questions

Impact of this question