1- A compound is 87.5% nitrogen and 12.5% hydrogen. The molecular weight of the compound is 32 g/mol. Describe the molecule referencing shape, and inter-molecular forces.
2- Gaseous butadiene, C4H6 , reacts with hydrogen gas in the presence of a platinum catalyst to form butane, C4H10 according to
C4H6 (g) +2 H2 → C4H10 (g)
A mixture of C4H6 and H2 known only to contain more H2 than C4H6 had a pressure of 74 mm Hg in an unknown volume. After the gas had been passed over the platinum catalyst, its pressure was 44 mm Hg in the same volume and temperature. What fraction of the molecules in the original mixture was butadiene?
3- A gaseous compound that only contains carbon, hydrogen, and sulfur is burned with oxygen under such conditions that the individual volume of the reactants and products can be measured at the same temperature and pressure. It is found that 3 volumes of the compound react with oxygen to yield 3 volumes of CO2 , 3 volumes of SO2 , and 6 volumes of water vapor. What volume of oxygen is required for combustion? What is the formula of the compound? Is this an empirical formula or a molecular formula?
4- A mixture of methane, CH4 and acetylene, C2H2 , occupied an unknown volume at a total pressure of 63 mm Hg. The sample was and the CO2 was collected. Its’ pressure was found to be 96 mm Hg in the same volume and temperature as the original mixture. What was the fraction of methane in the mixture?
5- A mixture of chromium and zinc metal weighing 0.362 gm was reacted with excess hydrochloric acid. After all of the metal reacted, 225 ml of dry hydrogen gas was collected at 27o C and 750 torr. Determine the mass percent of zinc in the mixture. Hint: Zinc reacts with HCl to produce zinc chloride and hydrogen gas. Chromium reacts with HCl to produce chromium(III)chloride and hydrogen gas.
1- The compound is composed of nitrogen (N) and hydrogen (H) with a molecular weight of 32 g/mol. To determine the molecule's formula, we need to find the ratio of the number of nitrogen and hydrogen atoms in the molecule. The atomic weight of nitrogen is 14 g/mol and hydrogen is 1 g/mol. From the given percentages, we can find that the compound contains 28 g/mol of nitrogen (87.5% of 32 g/mol) and 4 g/mol of hydrogen (12.5% of 32 g/mol). Thus, there are 28/14 = 2 nitrogen atoms and 4/1 = 4 hydrogen atoms in the molecule. So, the molecular formula is N2H4.
This molecule is known as hydrazine. The shape of the hydrazine molecule is bent, with each nitrogen atom bonded to two hydrogen atoms and a single bond between the nitrogen atoms. The molecule exhibits intermolecular forces like hydrogen bonding, dipole-dipole interactions and London dispersion forces.
2- Initially, the pressure of the mixture is 74 mm Hg, and after the reaction, it drops to 44 mm Hg. This means that the pressure difference, 30 mm Hg, corresponds to the amount of C4H6 and H2 consumed in the reaction. As the balanced equation shows that 1 mole of C4H6 reacts with 2 moles of H2, the pressure of consumed H2 is twice that of consumed C4H6. Let 'x' be the pressure of consumed C4H6, then the pressure of consumed H2 is 2x.
Thus, x + 2x = 30 mm Hg, which gives x = 10 mm Hg.
Therefore, the original pressure of butadiene C4H6 in the mixture is 10 mm Hg. Hence, the fraction of butadiene in the mixture is 10/74 ≈ 0.135 or 13.5%.
3- Let 'x' be the volume of O2 required for combustion. According to the problem, 3 volumes of the unknown compound react with x volumes of O2 to produce 3 volumes of CO2, 3 volumes of SO2, and 6 volumes of H2O.
Using the balanced equation, we can deduce the formula of the compound:
CnHmS + xO2 → nCO2 + CO2 + m/2 H2O + SO2
From the given information:
3n = 3 (CO2 volume)
3 = 3 (SO2 volume)
6m = 6 (H2O volume)
Thus, n = 1, m = 2. Therefore, the formula of the compound is CH2S.
Since the volumes of the reactants and products are equal, the unknown volume of O2 can be found:
3 + x = 3 + 3 + 6, which gives x = 3.
So, 3 volumes of O2 are required for combustion. The formula of the compound, CH2S, is the empirical formula.
4- The initial pressure of the mixture (methane and acetylene) was 63 mm Hg, and after the combustion (when only CO2 is left) it increased to 96 mm Hg. The difference in pressure, 33 mm Hg, accounts for the conversion of these hydrocarbons to CO2.
Let the pressure of methane in the initial mixture be ‘x’ mm Hg, then the pressure of acetylene will be (63 - x) mm Hg. In the balanced reactions, 1 volume of methane and 2.5 volumes of acetylene produce the same amount of CO2:
x = 2.5(63 – x)
This gives: x ≈ 42 mm Hg.
Hence the fraction of methane in the original mixture is 42/63 ≈ 0.67 or 67%.
5- We're given that 225 mL of dry hydrogen gas is collected at 27°C and 750 torr. First, use the ideal gas law to calculate the number of moles of hydrogen gas produced:
PV = nRT
Rearrange the equation to solve for n:
n = PV/RT
n = (750/760)*(225/1000)/((0.08206)*(27+273))
n ≈ 0.00984 moles
Since hydrogen gas is produced by the reaction of both zinc and chromium with HCl, let's determine the moles of zinc and chromium used:
Let x = moles of zinc and y = moles of chromium,
x + y = 0.00984
Since 1 mole of zinc reacts with HCl to produce 1 mole of hydrogen, and 1 mole of chromium reacts with 3 moles of HCl to produce 1 mole of hydrogen:
65x + 52y = 0.362 (total mass)
Now, we can solve for x and y:
x ≈ 0.00310 moles
y ≈ 0.00674 moles
Zinc mass percent = (0.00310 * 65) / 0.362 = 0.2002
Hence, the mass percent of zinc in the mixture is approximately 20.02%.
1- To describe the molecule, we first need to determine the empirical formula. We know that the compound is 87.5% nitrogen and 12.5% hydrogen by mass.
To find the empirical formula, you can assume a 100g sample of the compound. This means there would be 87.5g of nitrogen and 12.5g of hydrogen in that sample.
Next, we need to convert the masses of nitrogen and hydrogen to moles by dividing them by their respective molar masses. The molar mass of nitrogen is 14 g/mol and the molar mass of hydrogen is 1 g/mol.
Number of moles of nitrogen = 87.5g / 14 g/mol = 6.25 mol
Number of moles of hydrogen = 12.5g / 1 g/mol = 12.5 mol
Now, we need to find the simplest whole-number ratio of nitrogen to hydrogen. Divide both moles by the smallest number of moles (6.25 mol in this case):
Number of moles of nitrogen / 6.25 = 6.25 mol / 6.25 mol = 1 mol
Number of moles of hydrogen / 6.25 = 12.5 mol / 6.25 mol = 2 mol
The empirical formula is NH2, indicating that there is one nitrogen atom and two hydrogen atoms in each molecule.
To describe the shape, we need to know the Lewis structure or molecular geometry of the compound, which is not provided in the given information.