Question

The percent composition by mass of the phosphorus in calcium phosphate $C{a_3}{(P{O_4})_2}$ is $x$. The value $x$ is:

Answer 1

To answer this question, you should recall the method to find the percentage composition of an element in a compound. We shall assume 1 mole of the compound is present and use its molar mass and the molar mass of the elements present in it to find the percentage composition of each element.

Complete Step by step answer:
The percentage composition refers to the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100.
The molecular formula of calcium phosphate is $C{a_3}{(P{O_4})_2}$ ​:
Its molar mass is $3\left( {40} \right) + 2\left( {31} \right) + 8\left( {16} \right) = 310 {\text{g/mol}}.$
Percent of calcium =$\dfrac{{40 \times 3}}{{310}} \times 100 = 38.71\%$
Percent of phosphorus $= \dfrac{{31 \times 2}}{{310}} \times 100 = 20.0\%$
Percent of oxygen $= \dfrac{{16 \times 8}}{{310}} \times 100 = 41.29\%$

Hence, the value $x$ is $20.0\%$

Note: Other concentration terms used are:
Concentration in Parts Per Million (ppm) The parts of a component per million parts (${10^6}$) of the solution.
${\text{ppm(A) = }}\dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}} \times {10^6}$
$\% {\text{w/w}}$ is weight concentration of a solution: If a solution is labeled as $10\%$ glucose in water by mass, it refers to that 10g of glucose is dissolved in 90 g of water resulting in 100g of solution. $\% {\text{v/v}}$ is the volume concentration of a solution: it refers that if 50 mL of acetic acid is added to 50 mL of water, the acetic acid is labeled as $50\% {\text{v/v}}$. $\% {\text{w/v}}$ is the mass concentration of a solution: if x grams/ml of solute are present in solution it means x gram of solute X is dissolved in 100ml of solution. Suppose a solution contains solute A and solvent B, then its mass percentage is expressed as:
${\text{Mass \% of A = }}\dfrac{{{\text{Mass of component A in the solution}}}}{{{\text{Total mass of the solution}}}}{ \times 100}$
and Volume Percentage (V/V) can be written as ${\text{volume \% of A = }}\dfrac{{{\text{Volume of component A in the solution}}}}{{{\text{Total volume of the solution}}}}{\times 100}$
Molality (m): Molality establishes a relationship between moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent. The molality formula is as given- ${\text{Molality(m) = }}\dfrac{{{\text{Moles of solute}}}}{{{\text{Mass of solvent in kg}}}}$
Normality: It is defined as the number of gram equivalents of solute present in one litre of the solution.
Mole Fraction: It gives a unitless value and is defined as the ratio of moles of one component to the total moles present in the solution. Mole fraction = $\dfrac{{{X_{\text{A}}}}}{{{X_{\text{A}}} + {X_B}}}$ (from the above definition) where ${X_{\text{A}}}$ is no. of moles of glucose and ${X_{\text{B}}}$ is the no. of moles of solvent