Find the relative molecular mass (Mr) of NaOH, (NH4)2SO4, Na2CO3, Pb(NO3)2, and H2SO4

Answer

Using common relative atomic masses (Na 23, O 16, H 1, N 14, S 32, C 12, Pb 207), the relative molecular masses are: $M_r(\mathrm{NaOH})=40$, $M_r((\mathrm{NH_4})_2\mathrm{SO_4})=132$, $M_r(\mathrm{Na_2CO_3})=106$, $M_r(\mathrm{Pb(NO_3)_2})=331$, and $M_r(\mathrm{H_2SO_4})=98$.

Explanation

What you are calculating

Relative molecular mass, $M_r$, is the sum of the relative atomic masses of all atoms in the formula.

We will use these values: Na = 23, O = 16, H = 1, N = 14, S = 32, C = 12, Pb = 207.

$M_r$ of $\mathrm{NaOH}$

Count atoms: Na (1), O (1), H (1)

$$M_r = 23 + 16 + 1 = 40$$

$M_r$ of $\mathrm{(NH_4)_2SO_4}$

Expand the bracket: $(NH_4)_2$ means N (2) and H (8). Then add S (1) and O (4).

$$M_r = 2(14) + 8(1) + 32 + 4(16)$$ $$= 28 + 8 + 32 + 64 = 132$$

$M_r$ of $\mathrm{Na_2CO_3}$

Count atoms: Na (2), C (1), O (3)

$$M_r = 2(23) + 12 + 3(16)$$ $$= 46 + 12 + 48 = 106$$

$M_r$ of $\mathrm{Pb(NO_3)_2}$

First find one nitrate group:

$$M_r(\mathrm{NO_3}) = 14 + 3(16) = 14 + 48 = 62$$

There are 2 nitrates and 1 Pb:

$$M_r = 207 + 2(62) = 207 + 124 = 331$$

$M_r$ of $\mathrm{H_2SO_4}$

Count atoms: H (2), S (1), O (4)

$$M_r = 2(1) + 32 + 4(16)$$ $$= 2 + 32 + 64 = 98$$