Question

A 2.45-g sample of strontium completely reacts with oxygen to form 2.89 g of strontium oxide. Use this data to calculate the mass percent composition of strontium in strontium oxide.

Answer 1

Answer:

84.8% scrotium is present in 2.89g scrotium oxide

Explanation:

Mass percent of an element is the percentage product of the mass of the element divided by the mass of the compound formed by the interaction of the element with other elements.

Mathematically,

% of Sr element = mass of scrotium/ mass of scrotium oxide * 100%

% of Sr = 2.45g / 2.89g

% of Sr = 0.8477g of 100%

= 84.8% of scrotium.

Answer 2

Answer:

The % Mass Composition of Sr is 74.72≅  75%

Explanation:

The balanced equation:

Sr + O2 → SrO2

Molecular mass of SrO2 = 87.6 + (16 × 2) = 119.6 gmol-1

1 mole SrO2 = 119.6gmol-1

how many mole is in 2.5 g of SrO2 = 2.5÷119.6

= 0.0209 mol

The mole ration of the equation is:

1 mole SrO2 was produced by 1 Sr ion and 1 O2 ion

SrO2: Sr ratio is 1:1

⇒0.0209 mol SrO2 = 0.0209 mole of Sr

mass of Sr  = 0.0209 × 87.6gmol-1

= 1.83g

∴ & mass of Sr in sample = $\frac{1.83}{2.45}$ × 100

= 74.72≅  75%