learning goal: to use combustion analysis data to determine an empirical formula. a molecular formula…

learning goal: to use combustion analysis data to determine an empirical formula. a molecular formula expresses the number of each kind of atom in a molecule. for example, the molecular formula for propene, c3h6, indicates three carbon atoms and six hydrogen atoms per molecule. this also means that one mole of propene contains three moles of carbon and six moles of hydrogen. an empirical formula expresses the mole - ratio of the elements. the empirical formula for propene is ch2, indicating twice as much hydrogen as carbon. when analyzing unknown compounds in a lab, it is often possible to identify the mole ratios, and thus the empirical formula, but not the molecular formula. notice that the molecular mass of propene, 3(12)+6(1)=42 amu, is a multiple of the empirical formula mass, 1(12)+2(1)=14 amu. an unknown compound contains only carbon, hydrogen, and oxygen (cxhyoz). combustion of 6.50 g of this compound produced 9.53 g of carbon dioxide and 3.90 g of water. part a: how many moles of carbon, c, were in the original sample? express your answer to three significant figures and include the appropriate units. part b: how many moles of hydrogen, h, were in the original sample? express your answer to three significant figures and include the appropriate units.

learning goal: to use combustion analysis data to determine an empirical formula. a molecular formula expresses the number of each kind of atom in a molecule. for example, the molecular formula for propene, c3h6, indicates three carbon atoms and six hydrogen atoms per molecule. this also means that one mole of propene contains three moles of carbon and six moles of hydrogen. an empirical formula expresses the mole - ratio of the elements. the empirical formula for propene is ch2, indicating twice as much hydrogen as carbon. when analyzing unknown compounds in a lab, it is often possible to identify the mole ratios, and thus the empirical formula, but not the molecular formula. notice that the molecular mass of propene, 3(12)+6(1)=42 amu, is a multiple of the empirical formula mass, 1(12)+2(1)=14 amu. an unknown compound contains only carbon, hydrogen, and oxygen (cxhyoz). combustion of 6.50 g of this compound produced 9.53 g of carbon dioxide and 3.90 g of water. part a: how many moles of carbon, c, were in the original sample? express your answer to three significant figures and include the appropriate units. part b: how many moles of hydrogen, h, were in the original sample? express your answer to three significant figures and include the appropriate units.

Answer

Explanation:

Step1: Find moles of carbon dioxide

Use the molar - mass formula $n=\frac{m}{M}$, where $m$ is mass and $M$ is molar mass. The molar mass of $CO_2$ is $M_{CO_2}=12 + 2\times16=44\ g/mol$. Given $m_{CO_2}=9.53\ g$, then $n_{CO_2}=\frac{9.53\ g}{44\ g/mol}=0.2166\ mol$.

Step2: Determine moles of carbon

In one mole of $CO_2$, there is 1 mole of carbon. So, the moles of carbon in the original sample is equal to the moles of $CO_2$. Thus, $n_C = n_{CO_2}=0.217\ mol$ (rounded to three significant figures).

Step3: Find moles of water

The molar mass of $H_2O$ is $M_{H_2O}=2\times1 + 16 = 18\ g/mol$. Given $m_{H_2O}=3.90\ g$, then $n_{H_2O}=\frac{3.90\ g}{18\ g/mol}=0.2167\ mol$.

Step4: Determine moles of hydrogen

In one mole of $H_2O$, there are 2 moles of hydrogen. So, $n_H = 2\times n_{H_2O}=2\times0.2167\ mol = 0.433\ mol$ (rounded to three significant figures).

Answer:

Part A: Value: 0.217 Units: mol Part B: Value: 0.433 Units: mol