Question

MOLES AND MASS

Answer 1

Answer:

1) 0.428 mol

Here the compound that we have is $NaCl$.

We know that:

- Molar mass of sodium ($Na$): $22.990 g/mol$

- Molar mass of chlorine ($Cl$): $35.453 g/mol$

Since the ratio between $Na,Cl$ in the compound is 1:1, the molar mass of the $NaCl$ is:

$M_{NaCl}=M_{Na}+M_{Cl}=22.990+35.354=58.344 g/mol$

Here the mass of the compound is

$m=25 g$

Therefore, the number of moles is:

$n=\frac{m}{M}=\frac{25}{58.344}=0.428 mol$

2) 1.27 mol

The compound in this problem is

$H_2 SO_4$

The molar mass of each element is:

$M_H=1.0079 g/mol$ (hydrogen)

$M_S=32.065 g/mol$ (solphur)

$M_O=15.994 g/mol$ (oxygen)

In this compound, we have 2 atoms of hydrogen, 1 atom of solphur and 4 atoms of oxygen, therefore the total molar mass is:

$M_{H_2SO_4}=2M_H+M_S+4M_O=2(1.0079)+(32.065)+4(15.994)=98.057 g/mol$

Here the mass of the compound is

$m=125 g$

So the number of moles is

$n=\frac{m}{M}=\frac{125}{98.057}=1.27 mol$

3) 0.633 mol

The compound in this problem is

$KMnO_4$

The molar mass of each element is:

$M_K=39.0983 g/mol$ (potassium)

$M_{Mn}=54.9380 g/mol$ (manganese)

$M_O=15.994 g/mol$ (oxygen)

Here we have 1 atom of potassium, 1 atom of manganese and 4 atoms of oxygen, so the total molar mass of the compound is

$M=M_K+M_{Mn}+4M_O=(39.0983)+(54.9380)+4(15.994)=158.012 g/mol$

Here the mass of the compound is

$m=100 g$

So the number of moles is

$n=\frac{m}{M}=\frac{100}{158.012}=0.633 mol$

4) 0.993 mol

The compound in this problem is

$KCl$

The molar mass of each element is:

$M_K=39.0983 g/mol$ (potassium)

$M_{Cl}$ = $35.453 g/mol$ (chlorine)

So, the total molar mass of the compound is:

$M_{KCl}=M_K+M_{Cl}=39.0983+35.453=74.551 g/mol$

The mass of the compound here is

$m=74 g$

Therefore, the number of moles is

$n=\frac{m}{M}=\frac{74}{74.551}=0.993 mol$

5) 0.140 mol

The compound in this problem is

$CuSO_4\cdot 5H_2O$

The molar mass of each element is:

$M_{Cu}=63.546 g/mol$ (copper)

$M_S=32.065 g/mol$ (solphur)

$M_O=15.994 g/mol$ (oxygen)

$M_H=1.0079 g/mol$ (hydrogen)

Here we have 1 atom of copper, 1 of solphure, (4+5=9) of oxygen, and (5*2=10) atoms of hydrogen, so the total molar mass is

$M=M_{Cu}+M_S+9M_O+10M_H=\\=63.546+32.065+9(15.994)+10(1.0079)=249.636 g/mol$

Here the mass of the compound is

$m=35 g$

So, the number of moles is

$n=\frac{m}{M}=\frac{35}{249.636}=0.140 mol$