Several questions about stoichiometry

[Nyarka]

I need help for these questions... I do it tho, but some of them are wrong :/

1) The empirical formula of styrene is CH; the molar mass of styrene is 104.14 g/mol. What number of H atoms present in a 4.50g sample of styrene?

my answer is 8 because the ratio between C and H in styrene is 1:1. The way I get my answer is that (given molar mass)/(molar mass of CH) = 104.14/(12.01+1.008) = 7.94476655 ~8. Therefore there should be 8 H atoms present in a 4.50g sample of styrene.

[edited]: 8 x avogadro's number = no. of H atoms. = 4.8176 x 10^{24}

2) A compound contains only carbon, hydrogen, nitrogen and oxygen. Combustion of 0.157g of the compound produced 0.213g CO_2 and and 0.0310g H_2O. In another experiment, it is found that 0.103g of the compound produces 0.0230g NH_3. What is the empirical formula of the compound? Hint: Combustion involves reacting with excess O_2. Assume that all the carbon ends up in CO_2 and all the hydrogen ends up in H_2O. Also assume that all the nitrogen ends up in the NH_3 in the second experiment. (Type your answer using the format C2H5NO3 for C_2H_5_NO_3)

I don't really know how to solve it. I presume that the compound C_nH_nO_nN_n is X. I know that X + O_2 -> CO_2 + H_2O. molar mass of CO_2 is 44.01 and H_2O is 18.02. Since the given mass of CO_2 and H_2O is 0.213g and 0.0310g, the mole should be the given mass / molar mass, and I for CO_2 it has 0.213/44.01 = 4.839809134 x 10^{-3} mol, and H_2O has 0.0310/18.02 = 1.720310766 x 10^{-3} mol. I then divided the mole of CO_2 by mole of H_2O for the mole ratio and I got 2.813334213. So I can guess that the compound produce 3 mol of CO_2 when it produce 1 mol of H_2O.

Then after all I don't know how to do it. X -> NH_3 , actual mass of NH_3 = 0.0230g , molar mass of NH_3 = (14.01 + [1.008x3]), mol = 0.0230/(14.01 + 3.024) = 1.350240695 x 10^{-3} mol. Theoretically, if I want to know the molar mass of X, I can simply use it's given mass divided by mol. so 0.103/1.350240695 x 10^{-3} = 76.28269565.

Then I have no clue...

3) What number of Fe atoms and what amount (moles) of Fe atoms are in 250.0g of iron?

I know the answers of atoms but I got the wrong answer for moles.
atoms = (250.0/55.85) x 6.022x10^{23} = 2.69561325x10^{24} ~ 2.696x10^{24}

shouldn't moles of Fe = 250.0/55.85 = 4.476 mol? (at first I was 4.48 and I got it wrong, probably because of the sig. fig. ?)

4) What amount (moles) are represented by each of these samples?
- 18.0 mg NO_2

give mass / molar mass = mole, therefore (18.0/1000)/(32.00 + 14.01) = 3.912193002 x 10^{-4}, ~ 3.91x10^{-4} because the question used 3 sig. fig. for the mass of NO_2 sample.

5) Fungal lactase, a blue protein found in wood-rotting fungi, is 0.410% Cu by mass. If a fungal laccase molecule contains 4 copper atoms, what is the molar mass of fungal lactase?

The way I do it was, we know that there have 4 Cu atoms and by mass, it's only 0.410% out of the whole thing. So that, 0.410% = (63.55 x 4) = 254.2.

If mass of Fungal lactase is 1, Cu = 0.0041 out of 1

0.0041/1 = 254.2/ X,
X = 254.2/0.0041 = 62000. Molar mass of fungal lactase = 62000 = 620 x 10^{2} because the min sig. fig. number in the question is 3 (0.410%).

6) A sample of urea contains 1.233g N, 0.177g H, 0.528g C, and 0.704g O. What is the empirical formula of urea? (Type your answer using the format CO2 for CO_2)

Should I do it like this? total mass of urea = 1.233+0.177+0.528+0.704 = 2.642

No. of N in urea = 1.233/2.642 = 0.4666919 / 0.0669947 = 6.966101796 ~7
No. of H in urea = 0.177/2.642 = 0.0669947 / 0.0669947 = 1
No. of C in urea = 0.528/2.642 = 0.199848599 / 0.0669947 = 2.983050883 ~3
No. of O in urea = 0.704/2.642 = 0.266464799 / 0.0669947 = 3.977401182 ~4

Therefore, empirical formula of urea is C_3O_4N_7H

7) A compound contains only C, H and N. Combustion of 28.0mg of the compound produces 26.8mg CO_2 and 32.9mg H_2O. What is the empirical formula of the compound? (Type your answer using the format CO2 for CO_2)

X --> CO_2 + H_2O
mole of CO_2 = 26.8/(32 + 12.01) = 0.60895251
mole of H_2O = 32.9/18.01 = 1.826762909

we use the smaller one, therefore
mole of H_2O : CO_2 = 1.826762909/0.60895251 : 0.60895251/0.60895251 = 2.999844618:1 ~3:1

therefore
X --> 3CO_2 + H_2O

molar mass of X = 26.8/0.60895251 = 44.01

if we have 3 carbon and 2 hydrogen and N nitrogen in X, mass of C_3H_2 = (12.01x3)+(1.008x2)+(14.01xN) = 38.046 + 14.01xN > 44.01

can somebody help me? :/

8) Bornite (Cu_3FeS_3) is a copper ore used in the production of copper. When heated, the following reaction occurs.

2Cu_3FeS_3(s) + 7O_2(g) -> 6Cu(s) + 2FeO(s) + 6SO_2(g)

If 2.47 metric tons of bornite is reacted with excess O_2 and the process has an 82.9% yield of copper, what mass of copper is produced?

molar mass of bornite = (63.55x3) + 55.85 + (32.07x3) = 342.71

mole of bornite = 2.47/342.71 = 7.207259782 x 10^{-3}
mole of bornite:copper = 2:6
mole of copper = 7.207259782 x 10^{-3} x 3 = 0.021621779
molar mass of copper= 63.55
mass produced = 63.55 x 0.021621779 x 0.829 (yield) = 1.13909912 ~1.14g

I don't know if this is correct 'cause I'm not sure the "yield" part.

9) Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.

a) 0.450 mol of Ca(NO_3)_2 in 100.0mL of solution
Ca^{2+} is _____ M
NO_3^{-} is _____ M

b) 5.00g of NH_4Cl in 400.0mL of solution
NH_4^{+} is _____ M
Cl^{-} is _____ M

c) 1.00g K_3PO_4 in 500.0mL of solution
K^{+} is _____ M
PO_4^{3-} is _____ M

a) mol/volume = M, therefore M of Ca(NO_3)_2 is 0.450/(100/1000) = 4.5 x 10^{-6} M
because the mole ratio of Ca^{2+} : NO_3^{-} = 1:2
therefore
Ca^{2+} = 4.5 x 10^{-6} ~ 4.50 x 10^{-6}
NO_3^{-} = 4.5 x 10^{-6}x2 = 9.0 x 10^{-6} ~ 9.00 x 10^{-6}

b)molar mass of NH_4Cl = 14.01 + (1.008x4) + 35.45 = 53.492
mole of NH_4Cl = 5.00/53.492 = 0.093471921 mol
molarity of NH_4Cl = 0.093471921/(400/1000) = 2.336798026 x 10^{-7} M
because the mole ratio of NH_4^{+}:Cl^{-} is 1:1
therefore,
NH_4^{+} should have molarity 2.336798026 x 10^{-7} ~ 2.34 x 10^{-7}
Cl^{-} should also have molarity 2.336798026 x 10^{-7} ~ 2.34 x 10^{-7}

c)molar mass of K_3PO_4 = (39.10x3)+30.97+(16x4) = 212.27
mole of K_3PO_4 = 1/212.27 = 4.710981297 x 10^{-3} mol
molarity of K_3PO_4 = 4.710981297x10^{-3}/(500/1000) = 9.421962595 x 10^{-9}
because the mole ratio of K^{+}:PO_4^{3-} is 3:1
therefore
K^{+} should also have molarity 9.421962595 x 10^{-9} x 3 = 2.826588778 x 10^{-8} ~ 2.83 x 10^{-8}
PO_4^{3-} should also have molarity 9.421962595 x 10^{-9} ~ 9.42 x 10^{-9}

thanks for the help and reading

edit: those questions in "ultra" small sizes means solved :D

 

[symbolipoint]

Nvarka,

You seem to forget what "atom" and "mole" mean. They have each different meanings.

 

[symbolipoint]

If CH is the empirical formula, then the empirical grams per formula unit should be 13.02.

(4.5 grams CH)(1mole CH/13.02 grams CH)(1 mole H/1 mole CH)
= the number of MOLES of H

but you asked for how many ATOMS of H. You take it from here.

 

[Nyarka]

sorry man but I don't understand your explanation.

I don't think my answer is the mole of H instead of the number of atoms. If you read it carefully, what I did is divided the given molar mass by C_1H_1's molar mass. because

(CH)n <- the n is what I was looking for. so, I don't know. would be cool if you can tell the answer and explain it to me. thanks :]

 

[Nyarka]

no one want to help me out :/? I remember here used to have a few chemist always posting and solving people's question..

*pump*

 

[hage567]

So are you saying (3) is wrong? It looks OK to me. (4) looks good too. I haven't looked too closely at any of the others yet. I'm not a chemist, so it takes me a little longer to check this stuff.

 

[Nyarka]

basically all my answers for these questions are either totally wrong or not 100% correct.. that's why I'm looking for help here :x

 

[hage567]

1) The empirical formula of styrene is CH; the molar mass of styrene is 104.14 g/mol. What number of H atoms present in a 4.50g sample of styrene?

my answer is 8 because the ratio between C and H in styrene is 1:1. The way I get my answer is that (given molar mass)/(molar mass of CH) = 104.14/(12.01+1.008) = 7.94476655 ~8. Therefore there should be 8 H atoms present in a 4.50g sample of styrene.

[edited]: 8 x avogadro's number = no. of H atoms. = 4.8176 x 10^24

Find how many moles make up 4.5 g of CH (forget the business of dividing by (12.01 + 1.008)). How many molecules are in that number of moles? Use Avogadro's number here. Find that, and you will have your answer.

6) A sample of urea contains 1.233g N, 0.177g H, 0.528g C, and 0.704g O. What is the empirical formula of urea? (Type your answer using the format CO2 for CO_2)

Should I do it like this? total mass of urea = 1.233+0.177+0.528+0.704 = 2.642

No. of N in urea = 1.233/2.642 = 0.4666919 / 0.0669947 = 6.966101796 ~7
No. of H in urea = 0.177/2.642 = 0.0669947 / 0.0669947 = 1
No. of C in urea = 0.528/2.642 = 0.199848599 / 0.0669947 = 2.983050883 ~3
No. of O in urea = 0.704/2.642 = 0.266464799 / 0.0669947 = 3.977401182 ~4

Therefore, empirical formula of urea is C_3O_4N_7H

No. Find the number of moles of each element. Use the masses given. Don't worry about the total mass of the sample. To find the empirical formula, you divide each element (in moles) by the element that has the smallest number of moles. You are just taking the ratio with respect to the smallest number basically. This will give you the number of atoms of each element in the formula.