Question

Manganese makes up 1.3 × 10–4 percent by mass of the elements found in a normal healthy body. How many grams of manganese would be found in the body of a person weighing 196 lb? (2.205 lb = 1 kg) 56 g

Answer 1

Given the mass percentage of Manganese in a normal healthy body is $1.3*10^{-4}%$

It means that $1.3*10^{-4}$g Mn is present in 100 g of the body mass.

Given that the person weighs 196 lbs

Converting the person's mass to g: The conversion factors to be used are: 1 kg = 2.205 lb; 1 kg = 1000 g

$196lbs*\frac{1kg}{2.205lbs}*\frac{1000g}{1kg}= 8.89*10^{4}g$

Calculating the mass of Mn in the person's body:

$8.89*10^{4}g body*\frac{1.3*10^{-4}g Mn}{100 g body}=0.116g Mn$

Therefore, 0.116 g Mn is present in a person weighing 196 lbs.