Answer to Question #159626 in General Chemistry for Maddie Schuster

Question

1. A sample of a compound contains 88.8 grams of carbon and 11.2 grams of hydrogen. What is the empirical formula of this compound?

2. 19. How many grams of Al are in a sample of Al2(SO4)3 that contains 0.235 miles of O?

3. 2 Al(s)+3Fe(NO3)2(aq) -> 3F(s)+2Al(NO3)3(aq) If 5.89 grams of Al reacts with excess Fe(NO3)2, how many grams of Fe could be produced?

4. 3Na2So4+2AlCl3->6NaCl+Al2(SO4)3 5.67 grams of Na2SO4 reacted with 4.0 grams of AlCl3.

4a. Which reactant is the limiting and which one is the excess reactant and how many grams of Al2(SO4)3 are made?

4b. How many grams of the excess reactant will be leftover? 

5. A lab is performed where magnesium chloride is the desired product. Based on the data below, determine the empirical formula of magnesium chloride. Mass of empty beaker: 45.678 grams Mass of the beaker and magnesium: 46.002 grams Mass of the beaker and magnesium chloride: 47.420 grams

Answer 1

1. general mass:

$88,8+11,2=100$

find the percentages:

$\frac{88.8}{100}=0.888$ 88.8%

$\frac{11.2}{100}=0.112$ 11.2%

Divide the percentage by the atomic mass of the substance (ONE ATOM, so we divide hydrogen not by H2, but only by H):

$\frac{0.112}{1.008}=0.111$ (number of hydrogen atoms)

$\frac{0.888}{12}=0.074$ (number of carbon atoms)

Find the common divisor to get the integers, in this case the common divisor = 0.0185:

$\frac{0.111}{0.0185}=6$ (hydrogen atoms)

$\frac{0.074}{0.0185}=4$ (carbon atoms)

C4H6

2.19

moles of o = 0.235

The chemical formula of aluminium nitrate is 

Molar mass of O = 32 g/mol

Molar mass of aluminium = 27 g/mol

In 1 mole of Al2(SO4)3, 96 grams of O is combining with 54 grams of Al

So, 0.235 miles of O o will combine with

$\frac{54}{96}\times0.235=0.1322$

$m=n\times M=0.1322\times54=7.14$

3.2

$n{AL}=\frac{m}{M}=\frac{5.89}{54}=0.1091$

n(Fe}

$\frac{0.1091}{2}=\frac{x}{3}$

x=0.1637

$m (Fe)=n\times M=0.1637\times55.845=9.14$

4.

3Na2So4 + 2AlCl3-> 6NaCl + Al2 (SO4)

$n(Na2So4)=\frac{m}{M}=\frac{5.67}{142.042}=0.0399$

$n(AlCl3)=\frac{m}{M}=\frac{4}{133.3405}=0.0299$

4a. AlCl3 - limiting

$\frac{0.0299}{2}=\frac{x}{1}$

x=0.01499

$m =n\times M=0.01499\times150.0251=2.25$

4b.

$2.25\times\frac{1}{150.0251}\times\frac{3}{1}\times\frac{142.042}{3}=2.13$

5.67-2.13=3.54

5.

47.420-45.678=1.742 magnesium chloride.

46.002-45.678=0.324magnesium

$\frac{0.324}{1.742}=0.1860$ 18.60%

100%-18.60%=81.40 %

$\frac{0.1860}{24,3050}=0.00765$ (number of hydrogen atoms)

$\frac{0.8140}{35.457 }=0.0229$ (number of carbon atoms)

Find the common divisor to get the integers, in this case the common divisor = 0.0306

$\frac{0.00765}{0.0306}=0.25$ (hydrogen atoms)

$\frac{0.0229}{0.0306}=0.75$ (carbon atoms)

MgCl3