I need help for these questions... I do it tho, but some of them are wrong :/(adsbygoogle = window.adsbygoogle || []).push({});

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

therefore

[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

therefore

[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

therefore,

[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

therefore

[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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Several questions about stoichiometry

**Physics Forums | Science Articles, Homework Help, Discussion**