Question

a compound is found to consist of 94.08% oxygen and 5.92% hydrogen. what is the empirical formula of this compound? please enter your answers as numerical digits (i.e. 9). if any of your answers are 1, please write this even though the number 1 is generally assumed in chemical formulas. question 7 10 pts how many moles of nitrogen are contained in 39.08 g of (nh4)2co3? make sure to use the correct number of sig figs!

Explanation:

Step1: Determine moles of each element for empirical - formula problem

Assume 100 g of the compound. So, mass of oxygen ($m_O$) = 94.08 g and mass of hydrogen ($m_H$) = 5.92 g.
Molar - mass of oxygen ($M_O$)=16.00 g/mol and molar - mass of hydrogen ($M_H$)=1.01 g/mol.
Moles of oxygen ($n_O$)=$\frac{m_O}{M_O}=\frac{94.08\ g}{16.00\ g/mol}=5.88\ mol$.
Moles of hydrogen ($n_H$)=$\frac{m_H}{M_H}=\frac{5.92\ g}{1.01\ g/mol}=5.86\ mol$.

Step2: Find the mole - ratio

Divide each number of moles by the smaller number of moles. Here, the smaller number of moles is approximately 5.86 mol.
For hydrogen: $\frac{n_H}{n_{min}}=\frac{5.86\ mol}{5.86\ mol}=1$.
For oxygen: $\frac{n_O}{n_{min}}=\frac{5.88\ mol}{5.86\ mol}\approx1$.
So, the empirical formula is $H_1O_1$.

Step3: Determine moles of nitrogen for the second problem

Molar - mass of $(NH_4)_2CO_3$:
$M=(2\times(14.01 + 4\times1.01))+12.01+(3\times16.00)$
$M=(2\times(14.01 + 4.04))+12.01 + 48.00$
$M=(2\times18.05)+12.01 + 48.00$
$M = 36.10+12.01 + 48.00=96.11\ g/mol$.
In one mole of $(NH_4)_2CO_3$, there are 2 moles of nitrogen.
Moles of $(NH_4)_2CO_3$ in 39.08 g, $n=\frac{m}{M}=\frac{39.08\ g}{96.11\ g/mol}=0.4066\ mol$.
Moles of nitrogen, $n_N = 2\times n = 2\times0.4066\ mol = 0.813\ mol$ (to 3 sig - figs).

Answer:

For the empirical - formula problem: H 1, O 1
For the moles of nitrogen problem: 0.813