5) write the formula for copper (ii) sulfate pentahydrate.
a) cu₂so₃·h₅
b) cu₂s·h₂o
c) cus·5h₂o
d) (cuso₄)₅
e) cuso₄·5h₂o

6) determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
a) no
b) n₂o
c) no₂
d) n₂o₃
e) no₃
g4

7) determine the empirical formula for a compound that contains c, h and o. it contains 52.14% c and 34.73% o by mass.
a) c₂h₆o
b) cho
c) c₄h₁₃o₂
d) ch₄o₃
e) ch₃o

8) how many fe(ii) ions are there in 15.0 g of feso₄?
a) 1.64 × 10⁻²⁵ iron(ii) ions
b) 5.94 × 10²² iron(ii) ions
c) 6.10 × 10²⁴ iron(ii) ions
d) 1.37 × 10²⁷ iron(ii) ions

9) what is the stoichiometric coefficient for oxygen when the following equation is balanced using the lowest, whole - number coefficients?
______ ch₄o (l) + ______ o₂(g) → ______ co₂(g) + ______ h₂o(l)
a) 9
b) 7
c) 5
d) 3

Question

5) write the formula for copper (ii) sulfate pentahydrate.
a) cu₂so₃·h₅
b) cu₂s·h₂o
c) cus·5h₂o
d) (cuso₄)₅
e) cuso₄·5h₂o

6) determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
a) no
b) n₂o
c) no₂
d) n₂o₃
e) no₃
g4

7) determine the empirical formula for a compound that contains c, h and o. it contains 52.14% c and 34.73% o by mass.
a) c₂h₆o
b) cho
c) c₄h₁₃o₂
d) ch₄o₃
e) ch₃o

8) how many fe(ii) ions are there in 15.0 g of feso₄?
a) 1.64 × 10⁻²⁵ iron(ii) ions
b) 5.94 × 10²² iron(ii) ions
c) 6.10 × 10²⁴ iron(ii) ions
d) 1.37 × 10²⁷ iron(ii) ions

9) what is the stoichiometric coefficient for oxygen when the following equation is balanced using the lowest, whole - number coefficients?
______ ch₄o (l) + ______ o₂(g) → ______ co₂(g) + ______ h₂o(l)
a) 9
b) 7
c) 5
d) 3

Accepted Answer

### Question 5
# Explanation:
## Step1: Identify ions and hydrate
Copper (II) has charge $+2$, sulfate is $SO_4^{2-}$, so the anhydrous formula is $CuSO_4$. Pentahydrate means 5 water molecules, so the formula is $CuSO_4 \cdot 5H_2O$.
## Step2: Evaluate options
- A: Incorrect ions (sulfite, wrong hydrate).
- B: Incorrect ions (sulfide, wrong hydrate).
- C: Incorrect ion (sulfide).
- D: Incorrect (polymer-like, not hydrate).
- E: Matches $CuSO_4 \cdot 5H_2O$.
# Answer: E) $CuSO_4 \cdot 5H_2O$

### Question 6
# Explanation:
## Step1: Assume 100g sample
Mass of N: $36.86g$, O: $63.14g$.
## Step2: Calculate moles
Moles of N: $\frac{36.86g}{14.01g/mol} \approx 2.631mol$.
Moles of O: $\frac{63.14g}{16.00g/mol} \approx 3.946mol$.
## Step3: Find mole ratio
Divide by smallest (2.631):
N: $\frac{2.631}{2.631} = 1$, O: $\frac{3.946}{2.631} \approx 1.5$. Multiply by 2: $N_2O_3$? Wait, no—wait, 36.86/14 ≈ 2.63, 63.14/16≈3.94. 2.63/2.63=1, 3.94/2.63≈1.5. Wait, no, 36.86% N: 36.86/14 = 2.63, 63.14/16=3.94. 2.63:3.94 ≈ 2:3? Wait, no, 2.63*2=5.26, 3.94*2=7.88? No, wait, 36.86/14 = 2.63, 63.14/16=3.94. 2.63/2.63=1, 3.94/2.63≈1.5. So ratio N:O = 2:3? Wait, no, 1:1.5 is 2:3? Wait, no, 36.86% N: 36.86/14 = 2.63, 63.14/16=3.94. 2.63/2.63=1, 3.94/2.63≈1.5. So multiply by 2: N=2, O=3? Wait, but option D is $N_2O_3$, but wait, 36.86% N: 36.86/14 = 2.63, 63.14/16=3.94. 2.63/2.63=1, 3.94/2.63≈1.5. So N:O = 2:3? Wait, no, 1:1.5 is 2:3. But wait, 36.86% N: 36.86/14 = 2.63, 63.14/16=3.94. 2.63/2.63=1, 3.94/2.63≈1.5. So the empirical formula is $N_2O_3$? Wait, but let's check: 14*2 + 16*3 = 28 + 48 = 76. %N: 28/76 ≈ 36.84%, %O: 48/76≈63.16%, which matches. So answer D.
# Answer: D) $N_2O_3$

### Question 7
# Explanation:
## Step1: Assume 100g sample
Mass of C: $52.14g$, O: $34.73g$, H: $100 - 52.14 - 34.73 = 13.13g$.
## Step2: Calculate moles
Moles of C: $\frac{52.14g}{12.01g/mol} \approx 4.341mol$.
Moles of O: $\frac{34.73g}{16.00g/mol} \approx 2.171mol$.
Moles of H: $\frac{13.13g}{1.008g/mol} \approx 13.02mol$.
## Step3: Find mole ratio
Divide by smallest (2.171):
C: $\frac{4.341}{2.171} \approx 2$, O: $\frac{2.171}{2.171} = 1$, H: $\frac{13.02}{2.171} \approx 6$. So ratio C:H:O = 2:6:1, formula $C_2H_6O$ (matches option A). Wait, but let's check: 52.14% C: (24/46)*100≈52.17%, 34.73% O: (16/46)*100≈34.78%, H: (6/46)*100≈13.04%, which matches. So answer A.
# Answer: A) $C_2H_6O$

### Question 8
# Explanation:
## Step1: Molar mass of $FeSO_4$
Fe: 55.85, S:32.07, O:16.00*4. Molar mass: $55.85 + 32.07 + 64.00 = 151.92g/mol$.
## Step2: Moles of $FeSO_4$
$\frac{15.0g}{151.92g/mol} \approx 0.0987mol$.
Each $FeSO_4$ has 1 $Fe^{2+}$, so moles of $Fe^{2+}$ = 0.0987mol.
## Step3: Number of ions
$0.0987mol \times 6.022 \times 10^{23} ions/mol \approx 5.94 \times 10^{22} ions$ (matches option B).
# Answer: B) $5.94 \times 10^{22}$ iron(II) ions

### Question 9
# Explanation:
## Step1: Balance C, H, O
- C: $CH_4O$ has 1 C, so $CO_2$ coefficient 1.
- H: $CH_4O$ has 4 H, so $H_2O$ coefficient 2 (4 H / 2 per $H_2O$).
- O: Left: 1 (from $CH_4O$) + 2x (from $O_2$). Right: 2 (from $CO_2$) + 2 (from $H_2O$) = 4. So 1 + 2x = 4 → 2x=3 → x=1.5? Wait, no—wait, the formula is $CH_4O$ (methanol? Wait, $CH_4O$ is $CH_3OH$, molar formula $CH_4O$). Let's balance:
$aCH_4O + bO_2
ightarrow cCO_2 + dH_2O$
C: $a = c$
H: $4a = 2d \Rightarrow d = 2a$
O: $a + 2b = 2c + d$. Substitute $c=a$, $d=2a$:
$a + 2b = 2a + 2a \Rightarrow a + 2b = 4a \Rightarrow 2b = 3a \Rightarrow b = \frac{3a}{2}$. To make whole numbers, set $a=2$:
$2CH_4O + 3O_2
ightarrow 2CO_2 + 4H_2O$. Wait, no—wait, original equation: $CH_4O(l) + O_2(g)
ightarrow CO_2(g) + H_2O(l)$. Let's do it properly:

1. C: 1 on left, so $CO_2$ coefficient 1.
2. H: 4 on left, so $H_2O$ coefficient 2 (4 H / 2 per $H_2O$).
3. O: Left: 1 (from $CH_4O$) + 2b (from $O_2$). Right: 2 (from $CO_2$) + 2 (from $H_2O$) = 4. So 1 + 2b = 4 → 2b=3 → b=1.5. But we need whole numbers, so multiply all coefficients by 2:

$2CH_4O + 3O_2
ightarrow 2CO_2 + 4H_2O$. Wait, no—wait, the original equation is $CH_4O$ (1 C, 4 H, 1 O). So balancing:

- C: 1 → 1 (so $CO_2$ =1)
- H: 4 → 2 $H_2O$ (4 H)
- O: Left: 1 (from $CH_4O$) + 2x (from $O_2$). Right: 2 (from $CO_2$) + 2 (from $H_2O$) = 4. So 1 + 2x = 4 → x=1.5. But the question says "lowest whole-number coefficients", so multiply all by 2:

$2CH_4O + 3O_2
ightarrow 2CO_2 + 4H_2O$. Wait, but the options are A)9, B)7, C)5, D)3. Wait, no—wait, maybe I made a mistake. Wait, $CH_4O$ is $CH_3OH$ (methanol), formula $CH_4O$. Let's balance again:

$CH_4O + O_2
ightarrow CO_2 + H_2O$

C: 1=1
H: 4=2d → d=2
O: 1 + 2b = 2 + 2 → 1 + 2b = 4 → 2b=3 → b=1.5. So to get whole numbers, multiply by 2:

$2CH_4O + 3O_2
ightarrow 2CO_2 + 4H_2O$. So the coefficient for $O_2$ is 3 when $CH_4O$ is 2? Wait, no—wait, the original equation has $CH_4O$ as 1? Wait, no, the equation is:

$\_\_CH_4O(l) + \_\_O_2(g)
ightarrow \_\_CO_2(g) + \_\_H_2O(l)$

If we use $a=1$, then $b=1.5$, but we need whole numbers, so $a=2$, $b=3$, $c=2$, $d=4$. But the question is asking for the coefficient of $O_2$ when using lowest whole-number coefficients. Wait, but maybe I messed up the formula. Wait, $CH_4O$ is $CH_3OH$ (molar mass 32g/mol). Let's balance again:

1. C: 1 → 1 (CO₂=1)
2. H: 4 → 2 (H₂O=2)
3. O: 1 (from CH₄O) + 2x (O₂) = 2 (CO₂) + 2 (H₂O) → 1 + 2x = 4 → x=1.5. So to make x whole, multiply all by 2:

2 CH₄O + 3 O₂ → 2 CO₂ + 4 H₂O. So the coefficient of O₂ is 3 when CH₄O is 2? Wait, no—wait, the question's equation is written as $CH_4O$ (1 molecule), so the coefficients are:

2 CH₄O + 3 O₂ → 2 CO₂ + 4 H₂O. But the options are A)9, B)7, C)5, D)3. Wait, maybe the formula is $C_2H_4O$? No, the question says $CH_4O$. Wait, no—wait, maybe I made a mistake. Wait, let's check the options: the coefficient for O₂ is 3? Wait, no—wait, let's do it again. Let's take $a=1$ (CH₄O):

C: 1 → CO₂=1
H: 4 → H₂O=2
O: 1 (CH₄O) + 2b (O₂) = 2 (CO₂) + 2 (H₂O) → 1 + 2b = 4 → b=1.5. So the balanced equation with whole numbers is:

2 CH₄O + 3 O₂ → 2 CO₂ + 4 H₂O. So the coefficient of O₂ is 3 (when CH₄O is 2). But the question's equation is written as $CH_4O$ (1), so the coefficient for O₂ is 3/2? No, the question says "using the lowest, whole-number coefficients". So the coefficients are 2, 3, 2, 4. So the coefficient for O₂ is 3. So answer D.
# Answer: D) 3

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Explanation:

Question 5

Step1: Identify ions and hydrate

Step2: Evaluate options

Step1: Assume 100g sample

Step2: Calculate moles

Step3: Find mole ratio

Step1: Assume 100g sample

Step2: Calculate moles

Step3: Find mole ratio

Answer:

Question 6