Test corrections :) They're calculations/short answers. I'd just like something to compare answers to or help me get the full answer (all partial credits). Thanks!
#9. When water is electrolyzed it produces hydrogen and oxygen. If 10.0% of 150.0g of water is electrolyzed, wht is the final volume of gas if the temperature is 17.2 degrees C and the pressure is 98.9kPa?
#11. Caffeine has 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen and 16.49% oxygen. Its molar mass is 194.2g/mol. Its molecular formula is?
#14. The compound X2O is 63.7% X (mystery element). What is the identity of X?
#15. Carbon disulfide burns in oxygen. Complete combustion gives the reaction CS2 + 3O2 -> CO2 + 2SO2. [16.352 L SO2; 8.064 L CO2; Excess CS2] What was the pressure in the container after the reaction if the container is 3.86 L and the temperature was 392K?
I'm having trouble understanding what you want. Have you worked parts of the problems and you want us to work all of it in detail so you can compare your answers and see where you went wrong? I have a better idea. You show your work on each and I shall be happy to check the results and show corrections where needed.
Alright, sure.
#9.
2H2O -> 2H2 + O2
15g H2O/1 x 1mol/18.02 = 0.832 mol H2O
PV=nRT
(98.9kPa)(v)=(0.832mol)(8.31 kPaxL/molxK)(290.2K)
v=20.29L
#11.
12.01g/gC x 100=49.48%
24.27gC/1 x 1mol/12.01g=2.02mol C
1.01g/gH x 100=5.15%
19.61gH/1 x 1mol/1.01g=19.42mol H
14.01g/gN x 100=28.87%
48.53g/1 x 1mol/14.01g=3.46mol N
16g/gO x 100=16.49%
97.03g/1 x 1mol/16g=6.06mol O
2.02/2.02=1C
19.42/2.02=9H
3.46/2.02=2N
6.06/2.02=3O
N2CO3H9
#14.
O=36.3%
X=63.7%
16g/gO x 100=36.3%
44.08gO
44.08gO/1 x 1mol/16g= 2.755mol O
[That's as far as I got.]
#15.
(P)(3.86L)=(.39mol)(0.0821 atmxL/molxK)(392K)
P=3.25 atm
(P)(3.86)=(1.09mol)(0.0821)(392K)
P=9.09 atm
3.25+9.09=12.34 atm
#9. I get 20.32 using your numbers (I used 290.4 for T) but that should be rounded to 20.3. I think three s.f. are allowed (from the 10.0%). For oxygen, it will be half of that.
#11. First I think you have calculated g of the elements and not mols and second I think you've rounded too much. I do them this way.
Take a 100 g sample which give you
49.48 g C
5.15 g H
28.87 g N
16.49 g O
Then convert to mols.
49.48/12.01 = 4.11 mols C.
5.15/1 = 5.15 H
28.87/14.008 = 2.06
16.49/16 = 1.03
Divide all by the smallest and I get
3.99 C which rounds to 4.0o
5.00 H
2.00 N
1.00 O
empirical formula is C4H5N2O and the empircal mass is 97.1
Then empirical formula x Number = molar mass and solve for N. That is 2 and the molecular formula is C8H19N4O2.
#14 is #11 in reverse.
63.7 g X
36.3 g O
mols X = 63.7/atomic mass X = ?
mols O = 36.3/16 = 2.27
If the empirical formula is X2O that makes the 2.27 = 1; therefore X must be twice that or 4.54. That means
63.7/4.54 =14.04 which I would guess to be nitrogen. Check it out.
N2O = (14+14)/(14+14+16) = (28/44)*100 = 63.6% (You can get 63.7 if you use 14.04)
#15. I don't understand the 8.064 and 16.352
Thanks for posting your work.
#9. To find the final volume of gas produced when 10.0% of 150.0g of water is electrolyzed, we need to use the concept of stoichiometry and the Ideal Gas Law equation.
First, calculate the mass of water that is electrolyzed by multiplying the total mass of water by the percentage:
mass of water electrolyzed = 10.0% * 150.0g
Next, convert the mass of water electrolyzed to moles. Since the molar mass of water (H2O) is about 18.015 g/mol, divide the mass of water electrolyzed by the molar mass:
moles of water electrolyzed = mass of water electrolyzed / molar mass of water (H2O)
Use the balanced chemical equation for the electrolysis of water, which states that 2 moles of water produce 2 moles of hydrogen gas (H2) and 1 mole of oxygen gas (O2).
Since the stoichiometric ratio between water and the gases produced is 2:2:1, we can say that for every 2 moles of water electrolyzed, 2 moles of hydrogen gas and 1 mole of oxygen gas are produced.
Using the moles of water electrolyzed, calculate the moles of hydrogen gas produced:
moles of hydrogen gas = (moles of water electrolyzed / 2) * 2
The same logic applies to calculate the moles of oxygen gas:
moles of oxygen gas = moles of water electrolyzed / 2
Now, we can use the Ideal Gas Law equation to find the final volume of the gases produced. The Ideal Gas Law equation is:
PV = nRT
Where:
P = pressure (given: 98.9 kPa)
V = volume (unknown)
n = moles of gas (calculated above)
R = gas constant (0.0821 L·atm/(K·mol))
T = temperature in Kelvin (17.2°C + 273.15 = 290.35 K)
We can rearrange the Ideal Gas Law equation to solve for the volume:
V = (nRT) / P
Substitute the values into the equation and solve for the volume.
#11. To determine the molecular formula of caffeine given its percentage composition and molar mass, we need to calculate the empirical formula and then find the whole-number ratio of empirical formula units to the experimentally determined molar mass.
Start by assuming we have 100 grams of caffeine for ease of calculation. This assumption doesn't affect the final result.
Convert the percentage compositions of carbon, hydrogen, nitrogen, and oxygen into grams:
Carbon (C): 49.48 grams
Hydrogen (H): 5.15 grams
Nitrogen (N): 28.87 grams
Oxygen (O): 16.49 grams
Next, convert the masses of each element into moles by dividing by their respective atomic masses:
Carbon (C): moles = grams / atomic mass of carbon (12.01 g/mol)
Hydrogen (H): moles = grams / atomic mass of hydrogen (1.008 g/mol)
Nitrogen (N): moles = grams / atomic mass of nitrogen (14.01 g/mol)
Oxygen (O): moles = grams / atomic mass of oxygen (16.00 g/mol)
Divide each mole value by the smallest mole value obtained above to get whole-number ratios.
You can then multiply these ratios by the smallest whole number necessary to obtain whole-number ratios.
The final ratios give you the empirical formula of caffeine.
Finally, divide the experimentally determined molar mass of caffeine (194.2 g/mol) by the empirical formula's molar mass to find the multiplier.
Multiply the empirical formula by the multiplier to obtain the molecular formula of caffeine.
#14. To determine the identity of the element X in the compound X2O given its percentage composition, we need to calculate the molar mass of the compound and compare it to the molar masses of different elements.
Start by assuming we have 100 grams of X2O for ease of calculation. This assumption doesn't affect the final result.
Convert the percentage composition of X into grams:
X: 63.7 grams
Next, determine the molar mass of X2O by adding up the atomic masses of X and oxygen (O) using the periodic table.
Divide the molar mass of X2O by the molar mass of X to find the ratio between the molar masses.
Compare this ratio to the molar masses of different elements. Identify the element X in X2O based on the closest molar mass ratio.
#15. To find the pressure in the container after the combustion reaction of carbon disulfide (CS2) with oxygen (O2), we need to use the concept of stoichiometry and the Ideal Gas Law equation.
First, we need to calculate the moles of carbon disulfide (CS2) burned. Use the balanced chemical equation to determine the stoichiometry:
1 mole of CS2 reacts with 3 moles of O2 to produce 1 mole of CO2 and 2 moles of SO2.
Calculate the moles of carbon disulfide burned by dividing the given volume of carbon disulfide (CS2) by its molar volume at the given temperature and pressure.
Next, use the stoichiometry of the reaction to determine the moles of sulfur dioxide (SO2) produced. Since the stoichiometric ratio between carbon disulfide (CS2) and sulfur dioxide (SO2) is 1:2, multiply the moles of CS2 burned by 2 to find the moles of SO2 produced.
Now, use the Ideal Gas Law equation to find the pressure. The equation is: PV = nRT
Given:
- Volume (V) = 3.86 L
- Moles of sulfur dioxide (SO2) = calculated above
- Gas constant (R) = 0.0821 L·atm/(K·mol)
- Temperature (T) = 392 K
Rearrange the Ideal Gas Law equation to solve for pressure (P):
P = (nRT) / V
Substitute the values into the equation and solve for the pressure.