Question

What are the steps to solving these problems? I don’t need answers, just the steps on how to get the answers.

1. Find the molecular mass of Calcium Citrate.


4. What is the mass of 8.94 x 10^21 molecules of ethanol (C2H5OH)?


5. How many formula units are in 926 g of FeBr3?

Thanks!

Answer 1

1)Molar mass of calcium citrate is 498.46g/mol

4) Mass of
`8.94* 10^(21)` molecules of ethanol is
`68.29* 10^(-3)g`

5)926 gm of
`FeBr_(3)` has
`18.85* 10^(23)formula\,units`

Step-by-step explanation:

1)

The molecular formula of Calcium citrate-
`Ca_(3)(C_(6)H_(5)O_(7))_(2)`

Let's calculate the molar mass.

`3* 40g/mol+2((6* 12)+(5*1)+(7* 16))= 498.46g/mol`

Therefore, molar mass of calcium citrate is 498.46g/mol

4)

One mole of molecule has
`6.02* 10^(23)` molecules.

Molar mass of ethanol = 46.07 g/mol

`Mass=(8.94* 10^(21))* ((1mole)/(6.023* 10^(23)))*((46g)/(1mole))=68.29* 10^(-3)g`

Therefore, mass of
`8.94* 10^(21)` molecules of ethanol is
`68.29* 10^(-3)g`

5)

Molar mass of
`FeBr_(3)` = 295.56g/mol

One mole has
`6.02* 10^(23)` formula units.

Given mass = 926 g

`926 * (6.023* 10^(23)\,formula\,units)/(295.56)=18.85* 10^(23)formula\,units`

Therefore, 926 gm of
`FeBr_(3)` has
`18.85* 10^(23)formula\,units`