1. find the percent composition of each compound as given:
a. % o in feo
b. % cl in mgcl₂
c. % c in ch₄
d. % s in caso₄
e. % cl in hcl
2. a compound consisting of aluminum and chlorine weighs 17.82 grams. the aluminum in the compound weighs 3.70 grams.
a. what is the weight of the chlorine in the compound?
b. what is the experimental % al in the compound?
c. if the formula is alcl₃, what is the expected % al?
d. what is the percent error?
|expected - experimental| x 100 / expected

Question

1. find the percent composition of each compound as given:
    a. % o in feo
    b. % cl in mgcl₂
    c. % c in ch₄
    d. % s in caso₄
    e. % cl in hcl
2. a compound consisting of aluminum and chlorine weighs 17.82 grams. the aluminum in the compound weighs 3.70 grams.
    a. what is the weight of the chlorine in the compound?
    b. what is the experimental % al in the compound?
    c. if the formula is alcl₃, what is the expected % al?
    d. what is the percent error?
|expected - experimental| x 100 / expected

Accepted Answer

### Problem 1a: % O in FeO
# Explanation:
## Step1: Find molar mass of FeO
Molar mass of Fe ($M_{Fe}$) = $55.85\space g/mol$, molar mass of O ($M_{O}$) = $16.00\space g/mol$.
Molar mass of FeO ($M_{FeO}$)=$M_{Fe}+M_{O}=55.85 + 16.00=71.85\space g/mol$

## Step2: Calculate percent composition of O
Percent composition formula: $\%\text{O}=\frac{n\times M_{O}}{M_{FeO}}\times100$ (where $n = 1$ for O in FeO)
$\%\text{O}=\frac{1\times16.00}{71.85}\times 100=\frac{1600}{71.85}\approx22.3\%$

# Answer:
$22.3\%$

### Problem 1b: % Cl in $MgCl_{2}$
# Explanation:
## Step1: Find molar mass of $MgCl_{2}$
Molar mass of Mg ($M_{Mg}$)=$24.31\space g/mol$, molar mass of Cl ($M_{Cl}$)=$35.45\space g/mol$
Molar mass of $MgCl_{2}$ ($M_{MgCl_{2}}$)=$M_{Mg}+2\times M_{Cl}=24.31+2\times35.45 = 24.31 + 70.90=95.21\space g/mol$

## Step2: Calculate percent composition of Cl
Percent composition formula: $\%\text{Cl}=\frac{n\times M_{Cl}}{M_{MgCl_{2}}}\times100$ (where $n = 2$ for Cl in $MgCl_{2}$)
$\%\text{Cl}=\frac{2\times35.45}{95.21}\times100=\frac{70.90}{95.21}\times100\approx74.5\%$

# Answer:
$74.5\%$

### Problem 1c: % C in $CH_{4}$
# Explanation:
## Step1: Find molar mass of $CH_{4}$
Molar mass of C ($M_{C}$)=$12.01\space g/mol$, molar mass of H ($M_{H}$)=$1.01\space g/mol$
Molar mass of $CH_{4}$ ($M_{CH_{4}}$)=$M_{C}+4\times M_{H}=12.01 + 4\times1.01=12.01+4.04 = 16.05\space g/mol$

## Step2: Calculate percent composition of C
Percent composition formula: $\%\text{C}=\frac{n\times M_{C}}{M_{CH_{4}}}\times100$ (where $n = 1$ for C in $CH_{4}$)
$\%\text{C}=\frac{1\times12.01}{16.05}\times100=\frac{1201}{16.05}\approx74.8\%$

# Answer:
$74.8\%$

### Problem 1d: % S in $CaSO_{4}$
# Explanation:
## Step1: Find molar mass of $CaSO_{4}$
Molar mass of Ca ($M_{Ca}$)=$40.08\space g/mol$, molar mass of S ($M_{S}$)=$32.07\space g/mol$, molar mass of O ($M_{O}$)=$16.00\space g/mol$
Molar mass of $CaSO_{4}$ ($M_{CaSO_{4}}$)=$M_{Ca}+M_{S}+4\times M_{O}=40.08+32.07+4\times16.00=40.08 + 32.07+64.00=136.15\space g/mol$

## Step2: Calculate percent composition of S
Percent composition formula: $\%\text{S}=\frac{n\times M_{S}}{M_{CaSO_{4}}}\times100$ (where $n = 1$ for S in $CaSO_{4}$)
$\%\text{S}=\frac{1\times32.07}{136.15}\times100=\frac{3207}{136.15}\approx23.6\%$

# Answer:
$23.6\%$

### Problem 1e: % Cl in HCl
# Explanation:
## Step1: Find molar mass of HCl
Molar mass of H ($M_{H}$)=$1.01\space g/mol$, molar mass of Cl ($M_{Cl}$)=$35.45\space g/mol$
Molar mass of HCl ($M_{HCl}$)=$M_{H}+M_{Cl}=1.01+35.45 = 36.46\space g/mol$

## Step2: Calculate percent composition of Cl
Percent composition formula: $\%\text{Cl}=\frac{n\times M_{Cl}}{M_{HCl}}\times100$ (where $n = 1$ for Cl in HCl)
$\%\text{Cl}=\frac{1\times35.45}{36.46}\times100=\frac{3545}{36.46}\approx97.2\%$

# Answer:
$97.2\%$

### Problem 2a: Weight of chlorine in the compound
# Explanation:
## Step1: Use mass conservation
Let mass of compound $m_{compound}=17.82\space g$, mass of Al $m_{Al} = 3.70\space g$
Mass of Cl $m_{Cl}=m_{compound}-m_{Al}$

## Step2: Calculate mass of Cl
$m_{Cl}=17.82 - 3.70=14.12\space g$

# Answer:
$14.12\space g$

### Problem 2b: Experimental % Al
# Explanation:
## Step1: Use percent formula
Percent formula: $\%\text{Al (experimental)}=\frac{m_{Al}}{m_{compound}}\times100$

## Step2: Calculate experimental % Al
$\%\text{Al (experimental)}=\frac{3.70}{17.82}\times100=\frac{370}{17.82}\approx20.8\%$

# Answer:
$20.8\%$

### Problem 2c: Expected % Al in $AlCl_{3}$
# Explanation:
## Step1: Find molar mass of $AlCl_{3}$
Molar mass of Al ($M_{Al}$)=$26.98\space g/mol$, molar mass of Cl ($M_{Cl}$)=$35.45\space g/mol$
Molar mass of $AlCl_{3}$ ($M_{AlCl_{3}}$)=$M_{Al}+3\times M_{Cl}=26.98+3\times35.45=26.98 + 106.35=133.33\space g/mol$

## Step2: Calculate expected % Al
Percent composition formula: $\%\text{Al (expected)}=\frac{n\times M_{Al}}{M_{AlCl_{3}}}\times100$ (where $n = 1$ for Al in $AlCl_{3}$)
$\%\text{Al (expected)}=\frac{1\times26.98}{133.33}\times100=\frac{2698}{133.33}\approx20.2\%$

# Answer:
$20.2\%$

### Problem 2d: Percent error
# Explanation:
## Step1: Use percent - error formula
Percent - error formula: $\text{Percent Error}=\frac{\vert\text{Expected}-\text{Experimental}\vert}{\text{Expected}}\times100$
Expected $\%\text{Al}=20.2\%$, Experimental $\%\text{Al}=20.8\%$

## Step2: Calculate percent error
$\text{Percent Error}=\frac{\vert20.2 - 20.8\vert}{20.2}\times100=\frac{0.6}{20.2}\times100\approx3.0\%$

# Answer:
$3.0\%$

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QUESTION IMAGE

Question

Explanation:

Problem 1a: % O in FeO

Step1: Find molar mass of FeO

Step2: Calculate percent composition of O

Step1: Find molar mass of \(MgCl_{2}\)

Step2: Calculate percent composition of Cl

Step1: Find molar mass of \(CH_{4}\)

Step2: Calculate percent composition of C

Answer:

Problem 1b: % Cl in \(MgCl_{2}\)