Question 9c132

Apr 23, 2016

We just may apply expression for molarity.

Explanation:

According to molarity (or molar concentration) expression:

$M = \frac{n}{V} = \text{number of solute moles" / "litres of solution}$

We know that $V = 200 \text{ ml}$, now we need the number of moles for a 0.2 M solution:

0.2 " M" = {n ("C"_16 "H"_18 "N"_3 "SCl")}/{0.2 " l"} rightarrow

rightarrow n ("C"_16 "H"_18 "N"_3 "SCl") = 0.04 " mol"#

Now, let us calculate the mass of $\text{C"_16 "H"_18 "N"_3 "SCl}$ that we need to reach 0.04 mol. For this, we just take in account the molar mass of every elements:

• For carbon, C, we know that there are 16 atoms for every molecule of methylene blue. Its atomic mass is 12.01 amu (atomic mass units), so there are $12.01 \cdot 16 = \textcolor{b l u e}{192.16}$ amu of C per molecule. Analogously, we can say that there are 192.16 g of C per mol of methylene blue (atomic mass on amu and molar mass on g/mol are equivalent).

We just may apply this method to all the elements on methylene blue molecule:

• For hydrogen: $1.01 \cdot 18 = \textcolor{b l u e}{18.18}$ g/mol.
• For nitrogen: $14.01 \cdot 3 = \textcolor{b l u e}{42.03}$ g/mol.
• For sulphur: $32.07 \cdot 1 = \textcolor{b l u e}{32.07}$ g/mol.
• For chlorine: $35.43 \cdot 1 = \textcolor{b l u e}{35.43}$ g/mol.

The sum of all being:

$192.16 + 18.18 + 42.03 + 32.07 + 35.43 = \textcolor{g r e e n}{319.87}$ g/mol of molecule of methylene blue

If we need 0.04 mol of $\text{C"_16 "H"_18 "N"_3 "SCl}$, and each mole has a mass of 319.87 g, we will need:

$0.04 \cancel{\text{ mol" cdot {319.87 " g"}/{1 cancel " mol"} = color(red) 12.7948 " g}}$ of methylene blue