How do calculate the the following problems?

Note:
"v" signifies an upside down caret ^

v2 signifies the subscript of 2
v6 signifies the subscript of 6

Calculate each of the following:
A. number of Li atoms in 4.5 mol of Li
B. number of CO v2 molecules in 0.0180 mol of CO v2
C. Moles of Cu in 7.8 * 10^21 atoms of Cu
D. Moles of C v2 H v6 in 3.75 * 10^23 molecules of C v2 H v6

1 Answer
Oct 26, 2017

All of these can be solved by using unitary method.

A. 11 mol of Li = 6.022 * 10^23=6.0221023 atoms.
hence, 5.45.4 mol Li =5.4 * 6.022 * 10^23=5.46.0221023 atoms
= 3.252 * 10^24=3.2521024 atoms.

B. 11 mol of CO_2CO2 = 6.022 * 10^23=6.0221023 molecules.
Hence, 0.01800.0180 mol of CO_2CO2 = 0.0180 * 6.022 * 10^23=0.01806.0221023 molecules.

C. 6.022 * 10^236.0221023 atoms Cu = 11 mol Cu.
11 atom Cu = 1/(6.022*10^23)16.0221023 mol Cu.
hence, 7.8 * 10^217.81021 atoms of Cu = 7.8 * 10^21 * 1/(6.022*10^23)7.8102116.0221023 mol Cu.

D. 6.022 * 10^236.0221023 molecules of C_2H_6C2H6 = 11 mol C_2H_6C2H6.
11 molecule of C_2H_6C2H6 = 1/(6.022 * 10^23)16.0221023 mol C_2H_6C2H6.
Hence, 3.75 * 10^233.751023 molecules of C_2H_6C2H6 = 3.75 * 10^23 * 1/(6.022 * 10^23)3.75102316.0221023 mol C_2H_6C2H6.

I haven't bothered to do the complete calculations, you can do them yourself.