Question

Glycine is an amino acid made up of carbon, hydrogen, oxygen, and nitrogen atoms. Combustion of a 0.2036-g sample gives 132.9 mL of CO2 at 258C and 1.00 atm and 0.122 g of water. What are the percentages of carbon and hydrogen in glycine? Another sample of glycine weighing 0.2500 g is treated in such a way that all the nitrogen atoms are converted to N2(g). This gas has a volume of 40.8 mL at 258C and 1.00 atm. What is the percentage of nitrogen in glycine? What is the percentage of oxygen? What is the empirical formula of glycine?

Short answer

Question: Determine the empirical formula and percentages of each element in the chemical compound glycine based on the given information from combustion analysis and another experiment. Answer: The empirical formula of glycine is 𝐶₂𝐻₅𝑂₂𝑁. The percentages of each element in glycine are as follows: Carbon (32.06%), Hydrogen (6.68%), Oxygen (42.78%), and Nitrogen (18.48%).

Step-by-step solution

Determine the moles of CO2 and H2O produced in combustion.

We will use ideal gas law to determine the amount of CO2 produced in moles: \(PV=nRT \Rightarrow n = \frac{PV}{RT}.\) Given: \(P_{CO_2} = 1 \,atm\), \(V_{CO_2} = 132.9 \,mL \,(0.1329\,L)\), \(R=0.0821\,\frac{L\,atm}{K\,mol},\) and \(T = 25 + 273 = 298\,K\). Now, we find the moles of CO2: \(n_{CO_2}=\frac{(1\,atm)(0.1329\,L)}{(0.0821\,\frac{L\,atm}{K\,mol})(298\,K)} = 0.00544 \,mol.\) To find the moles of H2O produced, use its mass: \(0.122\,g\). The molar mass of water is 18.015 g/mol. Now, we find the moles of H2O: \(n_{H_2O} = \frac{0.122\,g}{18.015\,g/mol} = 0.00677\,mol.\)

Calculate the mass of carbon and hydrogen

One mole of CO2 contains one mole of Carbon. Therefore, moles of C = 0.00544 mol. Utilizing the molar mass of carbon (12.011 g/mol), we calculate the mass of carbon in the sample. mass of C = moles of C * molar mass of C = 0.00544 mol * 12.011 g/mol = 0.0653 g. Each mole of H2O contains two moles of Hydrogen. Therefore, moles of H = 2 * 0.00677 mol = 0.01354 mol. Utilizing the molar mass of hydrogen (1.008 g/mol), we calculate the mass of hydrogen in the sample. mass of H = moles of H * molar mass of H = 0.01354 mol * 1.008 g/mol = 0.0136 g.

Determine the percentages of Carbon and Hydrogen

We use the masses of carbon and hydrogen obtained in step 2 and divide them by the total sample mass to find the percentages of these elements in Glycine. Percentage of Carbon: \(\frac{0.0653\,g}{0.2036\,g} * 100\% = 32.06 \%\) Percentage of Hydrogen: \(\frac{0.0136\,g}{0.2036\,g} * 100\% = 6.68 \%\)

Determine moles of Nitrogen gas produced

Using the information given for the second experiment, we will determine the amount of Nitrogen (N2) in moles using the Ideal Gas Law: \(PV=nRT\Rightarrow n = \frac{PV}{RT}.\) With \(P_{N_2} = 1\,atm\), \(V_{N_2} = 40.8\,mL\,(0.0408\,L)\), \(R=0.0821\,\frac{L\,atm}{K\,mol},\) and \(T = 298\,K\). Now, we find the moles of N2: \(n_{N_2}=\frac{(1\,atm)(0.0408\,L)}{(0.0821\,\frac{L\,atm}{K\,mol})(298\,K)} = 0.00165 \,mol.\)

Calculate the mass of nitrogen in the sample

One mole of N2 contains two moles of nitrogen atoms. Therefore, moles of N = 2 * 0.00165 mol = 0.00330 mol. Utilizing the molar mass of nitrogen (14.01 g/mol), we calculate the mass of nitrogen in the sample. mass of N = moles of N * molar mass of N = 0.00330 mol * 14.01 g/mol = 0.0462 g.

Determine the percentage of Nitrogen

We divide the mass of nitrogen obtained in step 5 by the total sample mass (0.2500 g) to find the percentage of nitrogen in Glycine. Percentage of Nitrogen: \(\frac{0.0462\,g}{0.2500\,g} * 100\% = 18.48 \%\)

Calculate the percentage of Oxygen

Since we know the percentages of Carbon, Hydrogen, and Nitrogen, we can now calculate the percentage of Oxygen, as the sum of all these percentages should equal 100%. Percentage of Oxygen: \(100\% - 32.06\% - 6.68\% - 18.48\% = 42.78 \%\)

Determine the empirical formula of glycine

We will use the percentages of each element and assume 100 g of glycine to find moles of C, H, O, and N. Then, divide all moles by the smallest mole value to find the mole ratio and derive the empirical formula. Mole ratio C: \(\frac{32.06\,g}{12.011\,g/mol} = 2.67\,mol\) Mole ratio H: \(\frac{6.68\,g}{1.008\,g/mol} = 6.63\,mol\) Mole ratio O: \(\frac{42.78\,g}{16.00\,g/mol} = 2.68\,mol\) Mole ratio N: \(\frac{18.48\,g}{14.01\,g/mol} = 1.32\,mol\) Now we divide all mole ratios by the smallest value (1.32), we get: C = 2.03, H = 5.03, O = 2.03, and N = 1. The final empirical formula is 𝐶₂₉₋₈𝐻₅𝑂₂₃₋₈𝑁 (rounding to the nearest whole number gives 𝐶₂𝐻₅𝑂₂𝑁).