Question

Urea is a white crystalline solid used as a fertilizer, in the pharmaceutical industry, and in the manufacture of certain polymer resins. Analysis of urea reveals that, by mass, it is \(20.1 \%\) carbon, \(6.7 \%\) hydrogen, \(46.5 \%\) nitrogen, and the balance oxygen. (a) Find the empirical formula of urea. (b) A \(5.0-\mathrm{g} / \mathrm{L}\) solution of urea in water has an osmotic pressure of 2.04 atm, measured at \(25^{\circ} \mathrm{C}\). What are the molar mass and molecular formula of urea?

Short answer

The empirical formula is CH₄N₂O. The molar mass is approximately 60 g/mol. The molecular formula is CH₄N₂O.

Step-by-step solution

- Determine Mass of Each Element in 100g of Urea

Assume a 100g sample of urea. Based on the percentages: Carbon: 20.1g Hydrogen: 6.7g Nitrogen: 46.5g Oxygen: 100g - 20.1g - 6.7g - 46.5g = 26.7g.

- Convert Masses to Moles

Use atomic masses to convert grams to moles: Carbon: \(\frac{20.1g}{12.01g/mol} \approx 1.67 mol\) Hydrogen: \(\frac{6.7g}{1.01g/mol} \approx 6.63 mol\) Nitrogen: \(\frac{46.5g}{14.01g/mol} \approx 3.32 mol\) Oxygen: \(\frac{26.7g}{16.00g/mol} \approx 1.67 mol\)

- Determine the Simplest Whole Number Ratio

Divide each element's moles by the smallest number of moles calculated: Carbon: \(\frac{1.67 mol}{1.67 mol} = 1\) Hydrogen: \(\frac{6.63 mol}{1.67 mol} \approx 3.97 \approx 4\) Nitrogen: \(\frac{3.32 mol}{1.67 mol} \approx 1.99 \approx 2\) Oxygen: \(\frac{1.67 mol}{1.67 mol} = 1\). The empirical formula is CH₄N₂O.

- Determine Moles of Urea from Osmotic Pressure

Use the osmotic pressure formula \(\Pi = MRT\), where \(\Pi = 2.04 atm\), \(R = 0.0821 L·atm/(mol·K)\), and \(T = 298K\) (25°C in Kelvin). Solve for M (Molarity): \(2.04 atm = M \times 0.0821 \times 298\), so \(M = 0.0834 mol/L\).

- Calculate Molar Mass of Urea

The molarity of the solution (M) is the number of moles per liter. For a 5.0g/L solution: \(5.0g/L\div 0.0834 mol/L \approx 60 g/mol\).

- Determine Molecular Formula from Molar Mass

The empirical formula mass of CH₄N₂O: Carbon: \(12.01 g/mol\) Hydrogen: \(4 \times 1.01g/mol = 4.04 g/mol\) Nitrogen: \(2 \times 14.01g/mol = 28.02 g/mol\) Oxygen: \(16.00g/mol\) Total = \(12.01 + 4.04 + 28.02 + 16.00 = 60.07 g/mol\). Since the empirical formula mass (\(60.07 g/mol\)) matches the calculated molar mass, the molecular formula is the same as the empirical formula: CH₄N₂O.

Key concepts

urea composition analysis

Urea is often used as a fertilizer, in pharmaceuticals, and to make polymers. To find its empirical formula, let's start by analyzing its composition. The problem gives the percentage mass of each element in urea:

• Carbon: 20.1%
• Hydrogen: 6.7%
• Nitrogen: 46.5%
• Oxygen: balance

First, assume you have a 100g sample of urea, making the math straightforward:

• Carbon: 20.1g
• Hydrogen: 6.7g
• Nitrogen: 46.5g
• Oxygen: 100g - 20.1g - 6.7g - 46.5g = 26.7g

Next, convert these masses to moles using their atomic masses:

• Carbon: \(\frac{20.1g}{12.01g/mol} \approx 1.67 mol\)
• Hydrogen: \(\frac{6.7g}{1.01g/mol} \approx 6.63 mol\)
• Nitrogen: \(\frac{46.5g}{14.01g/mol} \approx 3.32 mol\)
• Oxygen: \(\frac{26.7g}{16.00g/mol} \approx 1.67 mol\)

To find the simplest whole number ratio, divide each mole value by the smallest number of moles:

• Carbon: \(\frac{1.67 mol}{1.67 mol}= 1\)
• Hydrogen: \(\frac{6.63 mol}{1.67 mol} \approx 3.97 \approx 4\)
• Nitrogen: \(\frac{3.32 mol}{1.67 mol} \approx 1.99 \approx 2\)
• Oxygen: \(\frac{1.67 mol}{1.67 mol}= 1\)

The empirical formula for urea is CH₄N₂O.

osmotic pressure calculation

Osmotic pressure can help determine the molar mass of substances in solution. For urea, we have a \(5.0\)-g/L solution with an osmotic pressure (\(\Pi\)) of 2.04 atm at \(25^{\circ} C\) or \(298K\).The osmotic pressure formula is: \[\Pi = MRT \] Here, \(R\) is the ideal gas constant (\(0.0821 L\cdot atm / (mol\cdot K)\)), and \(T\) is the temperature in Kelvin.Solve for Molarity (M): \(2.04 atm = M \times 0.0821 L\cdot atm/(mol\cdot K) \times 298K \)\[M = \frac{2.04}{(0.0821 \cdot 298)}\approx 0.0834 mol/L\] This molarity value means that there are \(0.0834 mol/L\) of urea in solution.

molar mass determination

Once the molarity is known, you can find the molar mass. The molar mass calculation uses the fact that molarity (M) is the number of moles per liter. We have a \(5.0 g/L\) solution of urea. From here, use the previously calculated molarity: \[5.0 g/L \div 0.0834 mol/L \approx 60 g/mol\] This calculation shows urea's molar mass to be approximately \(60 g/mol\).

molecular formula calculation

The molecular formula of urea will match its empirical formula if the empirical formula mass matches the molar mass. First, determine the mass of the empirical formula CH₄N₂O:

• Carbon: 12.01 g/mol
• Hydrogen: 4 \(times\) 1.01 g/mol = 4.04 g/mol
• Nitrogen: 2 \(times\) 14.01 g/mol = 28.02 g/mol
• Oxygen: 16.00 g/mol

Summing these up, the empirical formula mass is: \(12.01 + 4.04 + 28.02 + 16.00 = 60.07 g/mol\). Since the empirical formula's mass (\(60.07 g/mol\)) is very close to our calculated molar mass (\(60 g/mol\)), we conclude that the molecular formula of urea is the same as its empirical formula: CH₄N₂O.