# Question 6b970

Apr 8, 2015

The mass of the magnesium hydroxide present in the tablet is $\text{1 g}$.

So, you know that each tablet contains 500 mg of magnesium. However, this magnesium is actually a part of magnesium hydroxide, $M g {\left(O H\right)}_{2}$.

To see how much magnesium hydroxide you have in every tablet, you must first determine the percentage by mass magnesium has in $M g {\left(O H\right)}_{2}$. To do this, use their respective molar masses

(24.3050cancel("g/mol"))/(58.3197cancel("g/mol")) * 100 = "41.68%"

This means that you get 41.68 g of magnesium for every 100 g of $M g {\left(O H\right)}_{2}$. As a result, 500 mg of magnesium would correspond to

500 * 10^(-3)cancel("g Mg") * ("100 g "Mg(OH)_2)/(41.68cancel("g Mg")) = "1.199 g "Mg(OH)_2#

Rounded to one sig fig, the number of sig figs given for 500 mg, the answer will be

${m}_{M g {\left(O H\right)}_{2}} = \textcolor{g r e e n}{\text{1 g}}$