Question

How many moles of $Cr$ are there in 85g of $C{{r}_{2}}{{S}_{2}}$?
$\left( Cr=52,S=32 \right)$

Answer 1

Hint: Mole concepts are important while solving these problems. An atom will contribute some percentage of its mass and volume while the formation of its molecule.

Complete step by step answer:
The concepts of molar mass and number of moles are needed to solve the given illustration,
Molar mass-
Molar mass or molecular mass of a substance is the additive masses of the atoms present in the molecule multiplied by their respective subscripts.
As specified by Dalton, the molecular mass of one molecule is its mass expressed in unified atomic mass units. Mass of a molecule is based on 12 as the atomic weight of carbon-12.
For example,
Molar mass of ${{H}_{2}}O$ is calculated as,
Molar mass = $\left( 2\times 1 \right)+16$
                    = 18 g/mol
As,
Atomic mass of hydrogen = 1 gm/mol
Atomic mass of oxygen = 16 gm/mol.
Number of moles-
Number of moles of any molecule is the ratio of its given mass to the molar mass of the same.
It is the unitless quantity as it is the ratio of same quantities,

For example,
Number of moles of 36 gm ${{H}_{2}}O$,
Number of moles = $\dfrac{36}{18}$
                             = 2 moles.
Thus, according to the given data;
Molar mass of $C{{r}_{2}}{{S}_{2}}$ can be calculated as,
$C{{r}_{2}}{{S}_{2}}$ = $\left( 2\times 52 \right)+\left( 2\times 32 \right)$
         = 168 gm/mol
As,
Atomic mass of chromium = 52 gm/mol
Atomic mass of sulphur = 32 gm/mol
Thus, 1 mole of $C{{r}_{2}}{{S}_{2}}$ weighs = 168 gm/mol
168 gm $C{{r}_{2}}{{S}_{2}}$ contains 104 gm of $Cr$.
So, 85 gm of $C{{r}_{2}}{{S}_{2}}$ will contain,
$=\left( \dfrac{104}{168} \right)\times 85$ gm $Cr$
= 52.62 gm of $Cr$.

Thus, moles of $Cr$ in $C{{r}_{2}}{{S}_{2}}$ = $\dfrac{52.62gm}{104gm}$
                                             = 0.50 moles.
Therefore, moles of $Cr$ in 85g of $C{{r}_{2}}{{S}_{2}}$ are 0.50.

Note: Be specific while calculating the molecular masses as molecules and its subscripts are dignified by brackets.
Units and dimensions must be wisely considered while solving such problems.