Question

Calculate the molar mass for each of the following:

a.${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ , antiperspirant

b. ${\mathrm{KC}}_4{\mathrm{H}}_5{\mathrm{O}}_6$,cream of tartar

C. ${\mathrm{C}}_{16}{\mathrm{H}}_{19}{\mathrm{N}}_3{\mathrm{O}}_5\mathrm{S}$,amoxicillin, antibiotic

Short answer

a. The molar mass of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$islocalid="1653486831069" $342.02\mathrm{g}$

b. The molar mass of ${\mathrm{KC}}_4{\mathrm{H}}_5{\mathrm{O}}_6$islocalid="1653486840096" $187.07\mathrm{g}$

c. The molar mass of ${\mathrm{C}}_{16}{\mathrm{H}}_{19}{\mathrm{N}}_3{\mathrm{O}}_5\mathrm{S}$islocalid="1653486849731" $365.17\mathrm{g}$

Step-by-step solution

Part (a) Step 1 : Given information

We have to find the molar mass of${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$

Part (a) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{Al}$atom =localid="1653486888948" $26.98\mathrm{g}$

Molar mass of one $\mathrm{S}$atom =localid="1653486879575" $32.06\mathrm{g}$

Molar mass of one $\mathrm{O}$atom = localid="1653486870475" $15.99\mathrm{g}$

So, the molar mass of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$= $2\left(26.98\right)+3\left(32.06\right)\hspace{0.17em}+12\left(15.99\right)$=localid="1653486862184" $342.02\mathrm{g}$

Part (b) Step 1 : Given information

We have to find the molar mass of${\mathrm{KC}}_4{\mathrm{H}}_5{\mathrm{O}}_6$

Part (b) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{K}$atom = localid="1653486901851" $38.09\mathrm{g}$

Molar mass of one $\mathrm{C}$atom =localid="1653486910205" $12.01\mathrm{g}$

Molar mass of one $\mathrm{H}$atom =localid="1653486918422" $1.00\mathrm{g}$

Molar mass of one $\mathrm{O}$atom =localid="1653486928227" $15.99\mathrm{g}$

So, the molar mass of${\mathrm{KC}}_4{\mathrm{H}}_5{\mathrm{O}}_6$=$38.09+4\left(12.01\right)+5\left(1.00\right)+6\left(15.99\right)$=localid="1653486939028" $187.07\mathrm{g}$

Part (c) Step 1 : Given information

We have to find the molar mass of${\mathrm{C}}_{16}{\mathrm{H}}_{19}{\mathrm{N}}_3{\mathrm{O}}_5\mathrm{S}$

Part (c) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{C}$atom =localid="1653486951668" $12.01\mathrm{g}$

Molar mass of one$\mathrm{H}$atom =localid="1653486960525" $1.00\mathrm{g}$

Molar mass of one $\mathrm{N}$atom =localid="1653486969048" $14.00\mathrm{g}$

Molar mass of one $\mathrm{O}$atom =localid="1653486977271" $15.99\mathrm{g}$

Molar mass of one $\mathrm{S}$atom =localid="1653486985755" $32.06\mathrm{g}$

So, the molar mass of ${\mathrm{C}}_{16}{\mathrm{H}}_{19}{\mathrm{N}}_3{\mathrm{O}}_5\mathrm{S}$= $16\left(12.01\right)+19\left(1.00\right)+3\left(14.00\right)+5\left(15.99\right)+32.06$

=localid="1653487000547" $365.17\mathrm{g}$