QUESTION IMAGE
Question
practice
h₃po₄ h=3 p=4 o=4
pcl₅ p=1 cl=5
c₂h₆o c=2 h=6 o=
na₃po₄ na=3 p=4
fe₂o₃ fe=2 o=3
nh₃ n=1 h=3
so₂ s=1 o=2
fe(c₂h₃o₂)₃ fe=1 c=2 h=3 o=
ca₃(po₄)₂
hg₃(oh)₂
cu(no₃)₂
al₂(so₄)₃
2hg₃(po₄)₂
2hc₂h₃o₂ h=2+6=8 c=2=4 o=2=4
3ba₃(so₄)₃
2al(oh)₃
8fe(h₂o)₄(oh)₂ fe=8 h=64=86 o=32=48 o=
To solve for the number of oxygen atoms in each compound, we use the subscripts and coefficients (if present) to calculate the total number of O atoms. Let's go through a few examples:
Example 1: \( \boldsymbol{C_2H_6O} \)
- The subscript for O is 1 (since there's no number after O, it's 1).
- So, number of O atoms: \( 1 \).
Example 2: \( \boldsymbol{Fe(C_2H_3O_2)_3} \)
- Inside the parentheses: \( C_2H_3O_2 \) has \( O = 2 \).
- The coefficient outside the parentheses is 3, so multiply: \( 2 \times 3 = 6 \).
- So, number of O atoms: \( 6 \).
Example 3: \( \boldsymbol{2HC_2H_3O_2} \)
- Inside the compound: \( HC_2H_3O_2 \) has \( O = 2 \).
- The coefficient in front is 2, so multiply: \( 2 \times 2 = 4 \).
- So, number of O atoms: \( 4 \).
Example 4: \( \boldsymbol{8Fe(H_2O)_4(OH)_2} \)
- For \( (H_2O)_4 \): \( O = 1 \times 4 = 4 \) per formula unit.
- For \( (OH)_2 \): \( O = 1 \times 2 = 2 \) per formula unit.
- Total O per formula unit: \( 4 + 2 = 6 \).
- Multiply by the coefficient 8: \( 6 \times 8 = 48 \).
- So, number of O atoms: \( 48 \).
General Rule:
For a compound \( A_x(B_yO_z)_w \) (or with a coefficient \( c \) in front, \( cA_x(B_yO_z)_w \)):
- If O is in a group (e.g., \( O_z \) or \( (B_yO_z) \)), multiply the subscript of O by any subscripts outside the group and the coefficient in front.
Let’s apply this to all compounds:
| Compound | Calculation for O Atoms | O Atoms |
|---|---|---|
| \( PCl_5 \) | No O atoms (no O in formula) | 0 |
| \( C_2H_6O \) | \( O = 1 \) (subscript) | 1 |
| \( Na_3PO_4 \) | \( O = 4 \) (subscript) | 4 |
| \( Fe_2O_3 \) | \( O = 3 \) (subscript) | 3 |
| \( NH_3 \) | No O atoms (no O in formula) | 0 |
| \( SO_2 \) | \( O = 2 \) (subscript) | 2 |
| \( Fe(C_2H_3O_2)_3 \) | \( O = 2 \times 3 = 6 \) | 6 |
| \( Ca_3(PO_4)_2 \) | \( O = 4 \times 2 = 8 \) | 8 |
| \( Hg_3(OH)_2 \) | \( O = 1 \times 2 = 2 \) | 2 |
| \( Cu(NO_3)_2 \) | \( O = 3 \times 2 = 6 \) | 6 |
| \( Al_2(SO_4)_3 \) | \( O = 4 \times 3 = 12 \) | 12 |
| \( 2Hg_3(PO_4)_2 \) | \( O = 4 \times 2 = 8 \) per unit; \( 8 \times 2 = 16 \) | 16 |
| \( 2HC_2H_3O_2 \) | \( O = 2 \times 2 = 4 \) | 4 |
| \( 3Ba_3(SO_4)_3 \) | \( O = 4 \times 3 = 12 \) per unit; \( 12 \times 3 = 36 \) | 36 |
| \( 2Al(OH)_3 \) | \( O = 1 \times 3 = 3 \) per unit; \( 3 \times 2 = 6 \) | 6 |
| \( 8Fe(H_2O)_4(OH)_2 \) | \( O = (4 + 2) \times 8 = 48 \) | 48 |
Use this method to find O atoms for any compound by analyzing subscripts and coefficients!
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To solve for the number of oxygen atoms in each compound, we use the subscripts and coefficients (if present) to calculate the total number of O atoms. Let's go through a few examples:
Example 1: \( \boldsymbol{C_2H_6O} \)
- The subscript for O is 1 (since there's no number after O, it's 1).
- So, number of O atoms: \( 1 \).
Example 2: \( \boldsymbol{Fe(C_2H_3O_2)_3} \)
- Inside the parentheses: \( C_2H_3O_2 \) has \( O = 2 \).
- The coefficient outside the parentheses is 3, so multiply: \( 2 \times 3 = 6 \).
- So, number of O atoms: \( 6 \).
Example 3: \( \boldsymbol{2HC_2H_3O_2} \)
- Inside the compound: \( HC_2H_3O_2 \) has \( O = 2 \).
- The coefficient in front is 2, so multiply: \( 2 \times 2 = 4 \).
- So, number of O atoms: \( 4 \).
Example 4: \( \boldsymbol{8Fe(H_2O)_4(OH)_2} \)
- For \( (H_2O)_4 \): \( O = 1 \times 4 = 4 \) per formula unit.
- For \( (OH)_2 \): \( O = 1 \times 2 = 2 \) per formula unit.
- Total O per formula unit: \( 4 + 2 = 6 \).
- Multiply by the coefficient 8: \( 6 \times 8 = 48 \).
- So, number of O atoms: \( 48 \).
General Rule:
For a compound \( A_x(B_yO_z)_w \) (or with a coefficient \( c \) in front, \( cA_x(B_yO_z)_w \)):
- If O is in a group (e.g., \( O_z \) or \( (B_yO_z) \)), multiply the subscript of O by any subscripts outside the group and the coefficient in front.
Let’s apply this to all compounds:
| Compound | Calculation for O Atoms | O Atoms |
|---|---|---|
| \( PCl_5 \) | No O atoms (no O in formula) | 0 |
| \( C_2H_6O \) | \( O = 1 \) (subscript) | 1 |
| \( Na_3PO_4 \) | \( O = 4 \) (subscript) | 4 |
| \( Fe_2O_3 \) | \( O = 3 \) (subscript) | 3 |
| \( NH_3 \) | No O atoms (no O in formula) | 0 |
| \( SO_2 \) | \( O = 2 \) (subscript) | 2 |
| \( Fe(C_2H_3O_2)_3 \) | \( O = 2 \times 3 = 6 \) | 6 |
| \( Ca_3(PO_4)_2 \) | \( O = 4 \times 2 = 8 \) | 8 |
| \( Hg_3(OH)_2 \) | \( O = 1 \times 2 = 2 \) | 2 |
| \( Cu(NO_3)_2 \) | \( O = 3 \times 2 = 6 \) | 6 |
| \( Al_2(SO_4)_3 \) | \( O = 4 \times 3 = 12 \) | 12 |
| \( 2Hg_3(PO_4)_2 \) | \( O = 4 \times 2 = 8 \) per unit; \( 8 \times 2 = 16 \) | 16 |
| \( 2HC_2H_3O_2 \) | \( O = 2 \times 2 = 4 \) | 4 |
| \( 3Ba_3(SO_4)_3 \) | \( O = 4 \times 3 = 12 \) per unit; \( 12 \times 3 = 36 \) | 36 |
| \( 2Al(OH)_3 \) | \( O = 1 \times 3 = 3 \) per unit; \( 3 \times 2 = 6 \) | 6 |
| \( 8Fe(H_2O)_4(OH)_2 \) | \( O = (4 + 2) \times 8 = 48 \) | 48 |
Use this method to find O atoms for any compound by analyzing subscripts and coefficients!