Question

determine the number of each type of atom in these formulas:

1. hgco3

hg
c
o

1. 3 ca3(po4)2

ca
p
o

1. 2 nh4cn

n
h
c

1. 4 zn(no3)2

zn
n
o

1. mg(c2h3o2)2

mg
c
h
o

1. 3 cuso4·5h2o

cu
s
o
h

Explanation:

1. For \(HgCO_{3}\)

Step1: Count \(Hg\) atoms

There is no sub - script for \(Hg\), so the number of \(Hg\) atoms is \(1\).

Step2: Count \(C\) atoms

There is no sub - script for \(C\) (inside the formula \(HgCO_{3}\)), so the number of \(C\) atoms is \(1\).

Step3: Count \(O\) atoms

The sub - script for \(O\) is \(3\), so the number of \(O\) atoms is \(3\).

2. For \(3Ca_{3}(PO_{4})_{2}\)

Step1: Count \(Ca\) atoms

The coefficient is \(3\) and the sub - script for \(Ca\) is \(3\). Using the formula \(N = \text{coefficient}\times\text{sub - script}\), \(N_{Ca}=3\times3 = 9\).

Step2: Count \(P\) atoms

The coefficient is \(3\), and for \(P\) in \((PO_{4})_{2}\), the sub - script is \(1\) (implicit in \(PO_{4}\)). So \(N_{P}=3\times2\times1=6\).

Step3: Count \(O\) atoms

The coefficient is \(3\), and for \(O\) in \((PO_{4})_{2}\), the sub - script is \(4\). So \(N_{O}=3\times2\times4 = 24\).

3. For \(2NH_{4}CN\)

Step1: Count \(N\) atoms

In \(NH_{4}CN\), there are \(2\) \(N\) atoms (\(1\) from \(NH_{4}\) and \(1\) from \(CN\)). With a coefficient of \(2\), \(N_{N}=2\times2=4\).

Step2: Count \(H\) atoms

In \(NH_{4}CN\), the sub - script for \(H\) is \(4\). With a coefficient of \(2\), \(N_{H}=2\times4 = 8\).

Step3: Count \(C\) atoms

In \(NH_{4}CN\), there is \(1\) \(C\) atom. With a coefficient of \(2\), \(N_{C}=2\times1=2\).

4. For \(4Zn(NO_{3})_{2}\)

Step1: Count \(Zn\) atoms

The coefficient is \(4\) and the sub - script for \(Zn\) is \(1\). So \(N_{Zn}=4\times1 = 4\).

Step2: Count \(N\) atoms

The coefficient is \(4\), and for \(N\) in \((NO_{3})_{2}\), the sub - script is \(1\). So \(N_{N}=4\times2\times1=8\).

Step3: Count \(O\) atoms

The coefficient is \(4\), and for \(O\) in \((NO_{3})_{2}\), the sub - script is \(3\). So \(N_{O}=4\times2\times3=24\).

5. For \(Mg(C_{2}H_{3}O_{2})_{2}\)

Step1: Count \(Mg\) atoms

There is no coefficient (implicit \(1\)) and the sub - script for \(Mg\) is \(1\). So \(N_{Mg}=1\).

Step2: Count \(C\) atoms

The sub - script for \(C\) in \((C_{2}H_{3}O_{2})\) is \(2\), and the outer sub - script is \(2\). So \(N_{C}=2\times2=4\).

Step3: Count \(H\) atoms

The sub - script for \(H\) in \((C_{2}H_{3}O_{2})\) is \(3\), and the outer sub - script is \(2\). So \(N_{H}=3\times2=6\).

Step4: Count \(O\) atoms

The sub - script for \(O\) in \((C_{2}H_{3}O_{2})\) is \(2\), and the outer sub - script is \(2\). So \(N_{O}=2\times2=4\).

6. For \(3CuSO_{4}\cdot5H_{2}O\)

Step1: Count \(Cu\) atoms

The coefficient is \(3\) and the sub - script for \(Cu\) is \(1\). So \(N_{Cu}=3\times1=3\).

Step2: Count \(S\) atoms

The coefficient is \(3\) and the sub - script for \(S\) is \(1\). So \(N_{S}=3\times1 = 3\).

Step3: Count \(O\) atoms

For \(O\) in \(CuSO_{4}\): coefficient \(3\), sub - script \(4\). For \(O\) in \(H_{2}O\): coefficient \(3\) (from \(3CuSO_{4}\cdot5H_{2}O\)), sub - script \(1\) (in \(H_{2}O\)) and multiplier \(5\). \(N_{O}=(3\times4)+(3\times5\times1)=12 + 15=27\).

Step4: Count \(H\) atoms

The coefficient is \(3\) and for \(H\) in \(H_{2}O\) with sub - script \(2\). So \(N_{H}=3\times5\times2=30\).

Answer:

1. \(Hg:1\), \(C:1\), \(O:3\)
2. \(Ca:9\), \(P:6\), \(O:24\)
3. \(N:4\), \(H:8\), \(C:2\)
4. \(Zn:4\), \(N:8\), \(O:24\)
5. \(Mg:1\), \(C:4\), \(H:6\), \(O:4\)
6. \(Cu:3\), \(S:3\), \(O:27\), \(H:30\)