3. A liquid hydrocarbon (Cx Hy) is found to be 16.37% H by mass. A 1.158-g vaporized sample of the hydrocarbon has a 358 mL volume at 71.0 C and 749 mmHg. what is the molecular formula for this hydrocarbon?
4. A 1.00 g Sample of C7H603 (s), us completely burned in a bomb calorimeter assembly by a temperature increase of 5.80 C is noted. when the calorimeter absorbs 9.37 Kj of heat, its temperature increases by 2.48 C.
-what is the heat capicity of the calorimeter?
- what is the amount of heat given off when 1 mole of C7H6O3 (s) burned? (use qcal= C cal x delta T cal)
5. given that
N2 05 + H2O (l) ---> 2HNO3 (l) delta H = -73.7 KJ
2H2 (g) + 02 (G) ---> 2H2O (L) delta H = -571. 6 KJ
N2 (g) + 3O2 (g) + H2 (g) ---> 2HNO3 (L) delta H = -348.2 KJ
Calculate delta H final (standard enthalpy of formation) if dinitrogen pentoxide gas.
6. A gass effuse 1.55 times fatser than propane C3H8 (g) at the same temperature and pressure.
- is the gas heavier or lighter than propane?
- what is he molar mass of this gas?
3. To find the molecular formula of the hydrocarbon, we need to determine its empirical formula first.
Given that the hydrocarbon is 16.37% hydrogen by mass, we can assume a 100 g sample. Therefore, the hydrocarbon contains 16.37 g of hydrogen.
Next, we need to calculate the amount of carbon in the hydrocarbon. The remaining mass after subtracting the mass of hydrogen is 100 g - 16.37 g = 83.63 g. This is the mass of carbon.
Now, let's find the moles of hydrogen and carbon in the hydrocarbon. The molar mass of hydrogen is 1 g/mol, so the number of moles of hydrogen is 16.37 g / 1 g/mol = 16.37 mol. The molar mass of carbon is 12 g/mol, so the number of moles of carbon is 83.63 g / 12 g/mol = 6.969 mol.
Now, let's find the ratio of hydrogen to carbon in the empirical formula. To do this, we divide the number of moles of each element by the smallest number of moles (carbon, in this case).
The ratio becomes approximately 16.37 mol / 6.969 mol = 2.346 : 1.
Since we need to have whole numbers for the empirical formula, we can multiply this ratio by 3 to get a whole number ratio: 2.346 * 3 ≈ 7 : 3.
Therefore, the empirical formula of the hydrocarbon is C7H3.
4. To determine the heat capacity of the calorimeter, we can use the equation q = C × ΔT, where q is the heat absorbed, C is the heat capacity, and ΔT is the temperature change.
Given:
Mass of sample (C7H6O3) = 1.00 g
Temperature increase from burning = 5.80 °C
Heat absorbed by calorimeter (qcal) = 9.37 kJ
Temperature increase of calorimeter (ΔTcal) = 2.48 °C
Using q = C × ΔT, we can rearrange the equation to solve for C:
C = q / ΔT
Plugging in the values:
C = (9.37 kJ) / (2.48 °C) = 3.781 kJ/°C.
Therefore, the heat capacity of the calorimeter is 3.781 kJ/°C.
To calculate the amount of heat given off when 1 mole of C7H6O3 (s) is burned, we can use the given equation qcal = Ccal × ΔTcal.
Given:
Heat capacity of the calorimeter (Ccal) = 3.781 kJ/°C
Temperature increase of the calorimeter (ΔTcal) = 2.48 °C
Plugging in the values:
qcal = (3.781 kJ/°C) × (2.48 °C) = 9.368 kJ.
Therefore, the amount of heat given off when 1 mole of C7H6O3 (s) is burned is 9.368 kJ.
5. The enthalpy change for a reaction can be calculated using Hess's Law. In this case, we need to find the enthalpy change for the formation of dinitrogen pentoxide (N2O5) gas.
The given reactions are:
1) N2O5 (g) → 2HNO3 (l) ΔH = -73.7 kJ
2) 2H2 (g) + O2 (g) → 2H2O (l) ΔH = -571.6 kJ
3) N2 (g) + O2 (g) + H2 (g) → 2HNO3 (l) ΔH = -348.2 kJ
We can rearrange equation 3) by combining equation 1) and equation 2) to match the desired reaction:
N2 (g) + 3O2 (g) + H2 (g) → 2HNO3 (l)
By doing the necessary calculations:
2 x Equation 1) - Equation 2), we get:
2 x (-73.7 kJ) - (-571.6 kJ) = -147.4 kJ + 571.6 kJ = 424.2 kJ.
Therefore, the delta H final (standard enthalpy of formation) for dinitrogen pentoxide (N2O5) gas is 424.2 kJ.
6. To determine if the gas is heavier or lighter than propane (C3H8), we need to compare their molar masses.
Propane (C3H8) has a molar mass of:
3(12.01 g/mol) + 8(1.008 g/mol) = 44.097 g/mol.
Let's assume the molar mass of the unknown gas is M g/mol.
Given that the unknown gas effuses 1.55 times faster than propane at the same temperature and pressure, we can use Graham's law of effusion:
Rate1 / Rate2 = sqrt[(Molar Mass2) / (Molar Mass1)]
1.55 = sqrt[M / 44.097]
Squaring both sides of the equation:
2.4025 = M / 44.097
Solving for M:
M = 2.4025 x 44.097 = 105.306 g/mol
Therefore, the molar mass of the unknown gas is approximately 105.31 g/mol.
Since this molar mass is greater than that of propane, the unknown gas is heavier than propane.
3. To find the molecular formula of a hydrocarbon, we need to determine the empirical formula first. The empirical formula gives us the simplest whole number ratio of atoms present in a compound.
To find the empirical formula, we start by assuming we have 100g of the hydrocarbon. This assumption allows us to work directly with percent compositions as grams.
Given that the hydrocarbon is 16.37% hydrogen (H) by mass, we can assume we have 16.37g of H. We can calculate the number of moles of H by dividing the mass by its molar mass (1g/mol).
Molar mass of hydrogen (H) = 1g/mol
Number of moles of H = 16.37g / 1g/mol = 16.37 mol
Next, we need to determine the number of moles of carbon (C) in the hydrocarbon. To do this, we subtract the mass of hydrogen from the total mass of the compound.
Mass of hydrocarbon = 100g - 16.37g = 83.63g
Number of moles of C = 83.63g / molar mass of C
To determine the molar mass of carbon (C) and hydrogen (H), we need the atomic masses of carbon and hydrogen.
Molar mass of carbon (C) = 12.01g/mol
Molar mass of hydrogen (H) = 1.01g/mol
Calculating the molar mass of the hydrocarbon:
Molar mass of hydrocarbon (Cx Hy) = (molar mass of C × number of moles of C) + (molar mass of H × number of moles of H)
Now we know the molar mass of the hydrocarbon. The empirical formula tells us the simplest whole number ratio of carbon to hydrogen atoms in the compound. By dividing the molar mass of the hydrocarbon by the molar mass of the empirical formula, we can find this ratio.
Finally, to determine the molecular formula, we need additional experimental data such as the density or molecular weight of the compound.
4. To find the heat capacity of the calorimeter, we use the equation:
qcal = Ccal × ΔTcal
Where:
qcal is the heat absorbed by the calorimeter
Ccal is the heat capacity of the calorimeter
ΔTcal is the change in temperature of the calorimeter
Given that the calorimeter absorbs 9.37 KJ of heat and the temperature increases by 2.48 °C, we can substitute these values into the equation and solve for Ccal.
Ccal = qcal / ΔTcal
Substituting the values:
Ccal = 9.37 KJ / 2.48 °C
5. To calculate the delta H final (standard enthalpy of formation) of dinitrogen pentoxide (N2O5), we can use Hess's Law. Hess's Law states that the change in enthalpy for a chemical reaction is independent of the pathway taken.
We are given three chemical reactions with their respective enthalpy values. By manipulating these equations to match the desired final equation, we can determine the value of delta H final.
We want to match the final equation: N2O5(g)
Using the given reactions:
Reaction 1: N2(g) + 0.5O2(g) + H2(g) -> 2HNO3(l)
Given delta H = -348.2 kJ
Reaction 2: 2H2(g) + O2(g) -> 2H2O(l)
Given delta H = -571.6 kJ
Reaction 3: N2O5(g)
Given delta H = ? (to be determined)
By combining these reactions, we can cancel out certain species to achieve the desired final reaction:
N2(g) + O2(g) + 3H2(g) -> 2HNO3(l) + 2H2O(l)
Now we can find the delta H final:
delta H final = delta H3 + delta H2 - delta H1
delta H final = ? + (-571.6 kJ) - (-348.2 kJ)
Solving for delta H final will give us the standard enthalpy of formation for dinitrogen pentoxide (N2O5).
6. To determine if the gas is heavier or lighter than propane and calculate its molar mass:
We are given that the unknown gas effuses 1.55 times faster than propane (C3H8) at the same temperature and pressure.
The rate of effusion for a gas is inversely proportional to the square root of its molar mass. Therefore, if the unknown gas effuses faster than propane, it must have a smaller molar mass than propane.
We can set up the following ratio:
(Propane rate of effusion) / (Unknown gas rate of effusion) = sqrt(Propane molar mass) / sqrt(Unknown gas molar mass)
Since the unknown gas effuses 1.55 times faster, the ratio becomes:
1.55 = sqrt(molar mass of propane) / sqrt(molar mass of the unknown gas)
By squaring both sides and rearranging the equation, we can solve for the molar mass of the unknown gas:
Molar mass of the unknown gas = (molar mass of propane) / (1.55)^2