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Homework Help: Several questions about stoichiometry

  1. Mar 17, 2007 #1
    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 [tex]10^{24}[/tex]

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

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

    Then after all I don't know how to do it. [tex]X -> NH_3[/tex] , actual mass of [tex]NH_3[/tex] = 0.0230g , molar mass of [tex]NH_3[/tex] = (14.01 + [1.008x3]), mol = 0.0230/(14.01 + 3.024) = 1.350240695 x [tex]10^{-3}[/tex] 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 [tex]10^{-3}[/tex] = 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 6.022x[tex]10^{23}[/tex] = 2.69561325x[tex]10^{24}[/tex] ~ 2.696x[tex]10^{24}[/tex]

    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 [tex]NO_2[/tex]

    give mass / molar mass = mole, therefore (18.0/1000)/(32.00 + 14.01) = 3.912193002 x [tex]10^{-4}[/tex], ~ 3.91x[tex]10^{-4}[/tex] because the question used 3 sig. fig. for the mass of [tex]NO_2[/tex] 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 [tex]10^{2}[/tex] 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 [tex]CO_2[/tex])

    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 [tex]C_3O_4N_7H[/tex]

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

    [tex]X --> CO_2 + H_2O[/tex]
    mole of [tex]CO_2[/tex] = 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 [tex]H_2O : CO_2[/tex] = 1.826762909/0.60895251 : 0.60895251/0.60895251 = 2.999844618:1 ~3:1

    [tex]X --> 3CO_2 + H_2O[/tex]

    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 [tex]C_3H_2[/tex] = (12.01x3)+(1.008x2)+(14.01xN) = 38.046 + 14.01xN > 44.01

    can somebody help me? :/

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

    [tex]2Cu_3FeS_3(s) + 7O_2(g) -> 6Cu(s) + 2FeO(s) + 6SO_2(g)[/tex]

    If 2.47 metric tons of bornite is reacted with excess [tex]O_2[/tex] 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 [tex]10^{-3}[/tex]
    mole of bornite:copper = 2:6
    mole of copper = 7.207259782 x [tex]10^{-3}[/tex] 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 [tex]Ca(NO_3)_2[/tex] in 100.0mL of solution
    [tex]Ca^{2+}[/tex] is _____ M
    [tex]NO_3^{-}[/tex] is _____ M

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

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

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

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

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

    thanks for the help and reading

    edit: those questions in "ultra" small sizes means solved :D
    Last edited: Mar 18, 2007
  2. jcsd
  3. Mar 18, 2007 #2


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    You seem to forget what "atom" and "mole" mean. They have each different meanings.
  4. Mar 18, 2007 #3


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    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.
  5. Mar 18, 2007 #4
    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 [tex]C_1H_1[/tex]'s molar mass. because

    [tex](CH)n[/tex] <- 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 :]
    Last edited: Mar 18, 2007
  6. Mar 18, 2007 #5
    no one wanna help me out :/? I remember here used to have a few chemist always posting and solving people's question..

  7. Mar 18, 2007 #6


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    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.
    Last edited: Mar 18, 2007
  8. Mar 18, 2007 #7
    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
  9. Mar 19, 2007 #8


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    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.

    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.
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