Question

Part IV: Problem- solving questions.

Solve the following problems appropriately.
16. Calculate the molecular masses (in amu) of the following compounds:
sulfur dioxide (SO) a gas that is responsible for aid rain, and
caffeine (C,H, N,0,), a stimul ant present in tea, coffee, and cola
beverages.
17. How many moles of CO, are present in 176 g of CO,?
18. How many grams of oxygen can be prepared by the decomposition of 12
grams of mercury (0) oxide?
19. 25 g of NH, is mixed with 4 moles of 0, in the given reation:
4NH,(g) + 50,g) → 4NO(9) + 6H,O(|)
a. Which is the limiting reactant?
b. What mass of NO is formed?
C.
What mass of H,O is formed?
Consider the following reaction: 2N,O, → 4NO,g) + O,(g)
When 40 g of N,O, decomposes, 4.5 g of O, is formed. What is the percent
yield?
21. Two compounds have the same composition: 85.6 2% C and 14.38% H.
a. Obtain the empirical formula corresponding to this composition.
b. One of the compounds has a molecul ar mass of 28.03 amu; the other, of

Answer 1

This answer provides calculations for the molecular masses of sulfur dioxide and caffeine, the number of moles of CO2, the grams of oxygen produced from decomposing mercury oxide, the limiting reactant and masses of NO and H2O formed, the percent yield of O2 formed from N2O5, and the empirical formula and molecular mass of a compound with given composition.

Step-by-step explanation:

16. The molecular mass of sulfur dioxide (SO2) can be calculated by adding up the atomic masses of its constituent atoms. Sulfur has an atomic mass of 32.07 amu and oxygen has an atomic mass of 16.00 amu. Therefore, the molecular mass of SO2 is 32.07 + (16.00 * 2) = 64.07 amu.

For caffeine (C8H10N4O2), the atomic masses of carbon, hydrogen, nitrogen, and oxygen are 12.01 amu, 1.01 amu, 14.01 amu, and 16.00 amu respectively. Adding these up, the molecular mass of caffeine is 12.01 * 8 + 1.01 * 10 + 14.01 * 4 + 16.00 * 2 = 194.19 amu.

17. To determine the number of moles of CO2 present in 176 g, we need to divide the given mass by the molar mass of CO2. The molar mass of CO2 is 12.01 + 16.00 * 2 = 44.01 g/mol. Therefore, the number of moles of CO2 is 176 g / 44.01 g/mol = approximately 3.995 moles.

18. The balanced equation for the decomposition of mercury(II) oxide (HgO) is 2HgO -> 2Hg + O2. From the equation, we can see that 2 moles of HgO produce 1 mole of O2. So, to calculate the grams of oxygen produced from 12 grams of HgO, we need to convert the mass of HgO to moles and then use the mole ratio. The molar mass of HgO is 200.59 g/mol. Hence, 12 g of HgO is 12 g / 200.59 g/mol = 0.0598 moles. Since the mole ratio is 2:1, the mass of oxygen produced is 0.0598 moles * 32.00 g/mol = 1.91 g.

19. To determine the limiting reactant, we need to compare the number of moles of NH3 and O2 in the given reaction. From the balanced equation, we can see that 4 moles of NH3 react with 6 moles of O2. Using the molar masses of NH3 (17.03 g/mol) and O2 (32.00 g/mol), we can calculate the moles of each reactant. 25 g of NH3 is 25 g / 17.03 g/mol = 1.47 moles. 4 moles of O2 is 4 moles * (32.00 g/mol) / (2 * 16.00 g/mol) = 4 moles. Comparing the moles, we can see that NH3 is the limiting reactant because it has fewer moles. Therefore, NH3 is completely consumed, and there will be an excess of O2.

To calculate the mass of NO formed, we need to use the mole ratio from the balanced equation. From the equation, we can see that 4 moles of NH3 produce 4 moles of NO. Using the molar mass of NO (30.01 g/mol), we can calculate the mass of NO formed from 1.47 moles of NH3. The mass of NO is 1.47 moles * 30.01 g/mol = 44.11 g.

For the mass of H2O formed, we need to use the mole ratio from the balanced equation. From the equation, we can see that 4 moles of NH3 produce 6 moles of H2O. Using the molar mass of H2O (18.02 g/mol), we can calculate the mass of H2O formed from 1.47 moles of NH3. The mass of H2O is 1.47 moles * 6 moles * 18.02 g/mol = 158.77 g.

20. The given reaction shows that 2 moles of N2O5 produce 4 moles of NO2 and 1 mole of O2. If 40 g of N2O5 decomposes and 4.5 g of O2 is formed, we can calculate the percent yield. The molar mass of N2O5 is 108.02 g/mol. To calculate the theoretical yield of O2, we use the molar ratio from the balanced equation. The mole ratio of O2 to N2O5 is 1:1. So, if 40 g of N2O5 decomposes, the theoretical mass of O2 formed would be 40 g * (1 mole / 108.02 g/mol) * (1 mole O2 / 1 mole N2O5) * (32.00 g/mol) = 14.81 g. The percent yield can be calculated by dividing the actual mass of O2 formed (4.5 g) by the theoretical mass (14.81 g) and multiplying by 100%. The percent yield is 4.5 g / 14.81 g * 100% = 30.41%.

21. To find the empirical formula, we need to convert the percentages of carbon and hydrogen to moles. Assume that we have 100 g of the compound, then we have 85.62 g of carbon and 14.38 g of hydrogen. The molar mass of carbon is 12.01 g/mol, and the molar mass of hydrogen is 1.01 g/mol. Therefore, the number of moles of carbon is 85.62 g / 12.01 g/mol = 7.13 moles, and the number of moles of hydrogen is 14.38 g / 1.01 g/mol = 14.24 moles. To find the simplest whole number ratio, we divide the moles of each element by the smallest number of moles. In this case, we divide by 7.13 moles (the number of moles of carbon). The empirical formula is C1H2, which can be simplified to CH2.

b. One of the compounds has a molecular mass of 28.03 amu. Since the empirical formula is CH2, we assume that the molecular formula is C2H4. To calculate the molecular mass of C2H4, we multiply the subscripts by the atomic masses and add them up. The molecular mass of C2H4 is 2 * 12.01 amu + 4 * 1.01 amu = 28.06 amu.

Learn more about Molecular masses, Moles, Limiting reactant, Percent yield, Empirical formula