atoms to moles and more..

[ninobrn99]

I just can't seem to figure out how to work these freaking problems! I've read and read and it just isn't making any sense.

1. One million argon atoms is _______ mol of argon atoms.
2. There are _______ sulfer atoms in 25 molecules of C4H4S2.
3. The total number of atoms in .111 mol Fe(CO)3(PH3)2 is _______?
4. How many molecules of CH4 are in 48.2g of this compound?
5. Magnesium and nitrogen react in a combination reaction to produce magnesium nitride:
3Mg+N2=Mg3N2

In this particular experiment, a 9.27g sample of N2 reacts completely. The mass of Mg consumes is ______?

6. The combustion of propane (C3H8) produces CO2 and H2O:
C3H8+5O2 = 3CO2+4H2O

The reaction of 2.5 mol of O2 will produce ______mol of H2O.

I have the answers, but that doesn't help me for my test tomorrow. Can someone please explain how to do these?

[Russ]

Post the answers so I can check my method? I can't remember all this unfortunately and I don't want to give you incorrect information.

I'm thinking:
1) Divide 1 million by Avogadro's Number (6.02E23)
2) 25 molecules, each molecule has 2 atoms = 50
3) .111 mol = 6.02E23/9 atoms of a single element. Then multiply for all elements present
4) Convert 48.2g to mol (divide by formula mass) then to number of molecules (via Avogadro's number)
5) Convert N2 to mol (divide by formula mass) then use stoichiometry to find Mg, then convert back to mass
6) Stoichiometry (5 mol of O2 produces 4 mol of H2O. Therefore 2.5 mol produces half as much, 2 mol)

[ninobrn99]

Post the answers so I can check my method? I can't remember all this unfortunately and I don't want to give you incorrect information.

I'm thinking:
1) Divide 1 million by Avogadro's Number (6.02E23)
2) 25 molecules, each molecule has 2 atoms = 50
3) .111 mol = 6.02E23/9 atoms of a single element. Then multiply for all elements present
4) Convert 48.2g to mol (divide by formula mass) then to number of molecules (via Avogadro's number)
5) Convert N2 to mol (divide by formula mass) then use stoichiometry to find Mg, then convert back to mass
6) Stoichiometry (5 mol of O2 produces 4 mol of H2O. Therefore 2.5 mol produces half as much, 2 mol)
I did some searching online.
1) that is the correct method

2)I dont know what I was thinking. I reread it and saw that there were only 2 S atoms, so 2 * 25 is correct

3) still isnt making sense. It's 1.07 x 10^24, but I cant figure out how they got it. I did .0111*207.823*6.02*10^23 = 1.389176

4)I got 48.2*(1mol CH4/16.04 CH4)*(6.022*10^23/1 mol CH4)= 1.81*10^24

5)3Mg = 72.915; N2 = 28 }=100.9
9.27g/(2(14.007)) = .330913063
3 mol Mg/1 mol N2 = 3
3 * .330913063 = .992736189 * 24.3050(amu of Mg) = 24.13

6) 2.5 * (4 mol H20/5 mol O2)
=2.5 * .8H2O = 2 mol H2O

Can you shed a bit more light on 3?

[Russ]

3) still isnt making sense. It's 1.07 x 10^24, but I cant figure out how they got it. I did .0111*207.823*6.02*10^23 = 1.389176

Can you shed a bit more light on 3?

Why .0111?

In one mole there are 6.02E23 atoms. In .111 mole there are ~6.69E22 atoms (divided by 9).
That would be the answer if there was only one "lot" of atoms in your molecule.

However, we have the molecule Fe(CO)3(PH3)2
There's 1 Fe atom, 3 C atoms, 3 O atoms, 2 P atoms, 6 H atoms = 15 atoms overall.

Therefore 6.69E22 * 15 = 1.0033E24

That's close enough to the answer given for me to think it's a rounding error. Did you type out the original formula exactly?

[ninobrn99]

Why .0111?

In one mole there are 6.02E23 atoms. In .111 mole there are ~6.69E22 atoms (divided by 9).
That would be the answer if there was only one "lot" of atoms in your molecule.

However, we have the molecule Fe(CO)3(PH3)2
There's 1 Fe atom, 3 C atoms, 3 O atoms, 2 P atoms, 6 H atoms = 15 atoms overall.

Therefore 6.69E22 * 15 = 1.0033E24

That's close enough to the answer given for me to think it's a rounding error. Did you type out the original formula exactly?
I did. Here's the question again,
The total number of atoms in 0.111 mol of Fe(CO)3(PH3)2 is____________
a. 15
b. 1.07x10^24
c. 4.46x10^21
d. 1.67
e. 2.76x10^-24

[Russ]

I did. Here's the question again,
The total number of atoms in 0.111 mol of Fe(CO)3(PH3)2 is____________
a. 15
b. 1.07x10^24
c. 4.46x10^21
d. 1.67
e. 2.76x10^-24

If I got that question, I'd choose b).

It can't be d) or e) since you can't have "bits" of an atom (must be a whole number). a) is the amount of atoms in the formula and c) is probably an intermediate step.

[ninobrn99]

If I got that question, I'd choose b).

It can't be d) or e) since you can't have "bits" of an atom (must be a whole number). a) is the amount of atoms in the formula and c) is probably an intermediate step.
Works for me..thanks Russ.

[grgrsanjay]

yea got all right in my second attempt
ty for the questions