Question

In a clinical laboratory, a sample of urine containing \(0.120 \mathrm{~g}\) of urea \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\) (M. Wt. 60) was treated with excess of nitrous acid. The urea reacted according to the following equation \(\mathrm{NH}_{2} \mathrm{CONH}_{2}+2 \mathrm{HNO}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{~N}_{2}+3 \mathrm{H}_{2} \mathrm{O}\) The gas formed was passed through aqueous sodium hydroxide and final volume is measured at STP. What was the volume? (1) \(89.6 \mathrm{cc}\) (2) \(179.2 \mathrm{cc}\) (3) \(44.8 \mathrm{cc}\) (4) \(22.4 \mathrm{cc}\)

Short answer

The final volume of the gas is 89.6 cc.

Step-by-step solution

Determine the moles of urea

First, find the number of moles of urea in the sample. Use the formula: \text{moles} = \frac{\text{mass}}{\text{molar mass}}. Given the mass of urea is 0.120 g and the molar mass is 60 g/mol, calculate the moles of urea: \[ \text{moles of urea} = \frac{0.120 \text{ g}}{60 \text{ g/mol}} = 0.002 \text{ moles} \]

Use stoichiometry to find moles of gas formed

According to the balanced chemical equation: \[ \text{NH}_2 \text{CONH}_2 + 2 \text{HNO}_2 \rightarrow \text{CO}_2 + 2 \text{N}_2 + 3 \text{H}_2 \text{O} \]. From this, 1 mole of urea produces 2 moles of nitrogen gas (\(\text{N}_2\)). Hence, the moles of \(\text{N}_2\) produced are: \[ 0.002 \text{ moles urea} \times 2 = 0.004 \text{ moles} \text{ N}_2 \]

Convert moles of gas to volume at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies a volume of 22.4 liters (or 22,400 cubic centimeters). Calculate the volume of \(\text{N}_2\) produced: \[ \text{Volume} = 0.004 \text{ moles} \times 22,400 \text{ cc/mole} = 89.6 \text{ cc} \]

Conclusion

The final volume of the gas measured at STP is 89.6 cc.

Key concepts

Moles Calculation

Calculating moles is the foundation of solving stoichiometry problems. Knowing how to determine the number of moles of a substance is crucial. Moles represent the amount of substance and can be calculated using the formula:

• moles = \( \frac{mass}{molar \; mass} \)

For the given exercise, the mass of urea is 0.120 grams and the molar mass is 60 grams per mole. Plugging these values into the formula gives the moles of urea: \[ \text{moles of urea} = \frac{0.120 \; \text{g}}{60 \; \text{g/mol}} = 0.002 \; \text{moles} \]So, the sample contains 0.002 moles of urea. This calculation is the first step in determining how much product will be formed in a reaction.

Balanced Chemical Equations

A balanced chemical equation is essential for stoichiometry. It shows the relationship between reactants and products. Each side of the equation must have the same number of atoms for each element. This is necessary to obey the law of conservation of mass. In the exercise, the balanced equation is: \[ \text{NH}_2 \text{CONH}_2 + 2 \text{HNO}_2 \rightarrow \text{CO}_2 + 2 \text{N}_2 + 3 \text{H}_2 \text{O} \]From this, we see that one mole of urea produces two moles of nitrogen gas (\( \text{N}_2 \)). Since we have calculated that we have 0.002 moles of urea, we use stoichiometry to find the moles of nitrogen gas formed: \[ 0.002 \; \text{moles urea} \times 2 = 0.004 \; \text{moles} \; \text{N}_2 \]Thus, the sample produces 0.004 moles of nitrogen gas.

Gas Volume at STP

The concept of gas volumes at Standard Temperature and Pressure (STP) is vital in chemistry. STP conditions are 0°C (273.15 K) and 1 atmospheric pressure. At these conditions, one mole of any ideal gas occupies 22.4 liters (or 22,400 cubic centimeters). This value is a key component in converting moles of gas into volume. In the exercise, we previously calculated that 0.004 moles of nitrogen gas are produced. To find the volume, we use the relationship: \[ \text{Volume at STP} = \text{moles} \times 22,400 \; \text{cc/mole} \]Plugging in the moles of nitrogen gas: \[ \text{Volume} = 0.004 \; \text{moles} \times 22,400 \; \text{cc/mole} = 89.6 \; \text{cc} \]So, the final volume of nitrogen gas produced at STP is 89.6 cubic centimeters. This precise step demonstrates the practical application of these concepts in determining gas volumes through stoichiometry.