Question

Calculate the molar mass for each of the following :

a. ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_4$ b. ${\mathrm{Ga}}_2{\left({\mathrm{CO}}_3\right)}_3$ c. ${\mathrm{KBrO}}_4$

Short answer

a. The molar mass of ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_4$islocalid="1653486224646" $120\mathrm{g}$

b. The molar mass of ${\mathrm{Ga}}_2{\left({\mathrm{CO}}_3\right)}_3$islocalid="1653486235157" $319.38\mathrm{g}$

c. The molar mass of ${\mathrm{KBrO}}_4$islocalid="1653486250834" $240.38\mathrm{g}$

Step-by-step solution

Part (a) Step 1 : Given information 

We have to calculate the molar mass of ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_4$

Part (a) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{C}$atom =localid="1653486264374" $12.01\mathrm{g}$

Molar mass of one $\mathrm{H}$atom =localid="1653486272836" $1.00\mathrm{g}$

Molar mass of one $\mathrm{O}$ atom = localid="1653486282065" $15.99\mathrm{g}$

So, molar mass of localid="1652518930399" ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_4$=$4\left(12.01\right)+8\left(1.00\right)+4\left(15.99\right)$=localid="1653486292791" $120\mathrm{g}$

Part (b) Step 1 : Given information

We have to calculate the molar mass of ${\mathrm{Ga}}_2{\left({\mathrm{CO}}_3\right)}_3$

Part (b) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{Ga}$atom =localid="1653486305517" $69.72\mathrm{g}$

Molar mass of one $\mathrm{C}$atom =localid="1653486315710" $12.01\mathrm{g}$

Molar mass of one$\mathrm{O}$atom =localid="1653486324766" $15.99\mathrm{g}$

So, molar mass of ${\mathrm{Ga}}_2{\left({\mathrm{CO}}_3\right)}_3$=$2\left(69.72\right)+3\left(12.01\right)+9\left(15.99\right)$=localid="1653486333806" $319.38\mathrm{g}$

Part (c) Step 1 : Given information

We have to calculate the molar mass of ${\mathrm{KBrO}}_4$

Part (c) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{K}$atom =localid="1653486346463" $39.09\mathrm{g}$

Molar mass of one $\mathrm{Br}$ atom =localid="1653486357129" $137.33\mathrm{g}$

Molar mass of one$\mathrm{O}$atom =localid="1653486366275" $15.99\mathrm{g}$

So, molar mass of ${\mathrm{KBrO}}_4$=$39.09+137.33+4\left(15.99\right)$=localid="1653486375410" $240.38\mathrm{g}$