Question

(Need all boxes answered)

Bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. Its formula is NaHCO3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. You may use whole numbers for the masses.

The mass of one mole of NaHCO3 is:

Answer 1

Here's what I get.

Step-by-step explanation:

`\text{The mass of one mole of NaHCO$_(3)$ is:}\\\\\text{$\boxed{23}$ g Na +$\boxed{1}$ g H + $\boxed{12}$ g C + $\boxed{48}$ g O = $\boxed{84}$ g}`

`\text{mass $\%$ Na = $\boxed{23}$ g $/$ $\boxed{84}$ g $*$ 100 = $\boxed{27.38}$ $\%$}\\\\\text{mass $\%$ H = $ \quad \boxed{1}$ g $/$ $\boxed{84}$ g $*$ 100 = $\boxed{1.19}$ $\%$}\\\\\text{mass $\%$ C = $\:\: \boxed{12}$ g $/$ $\boxed{84}$ g $*$ 100 = $\boxed{14.29}$ $\%$}\\\\\text{mass $\%$ O = $\:\: \boxed{48}$ g $/$ $\boxed{84}$ g $*$ 100 = $\boxed{57.14}$ $\%$}`

`\boxed{27.38} + \boxed{1.19} + \boxed{14.29} + \boxed{ 57.14} = \boxed{100.00}`

Answer 2

Answer & Explanation:

• If we have 1.0 mol of NaHCO₃:

∴ The mass of 1.0 mol of NaHCO₃ = (no. of moles of NaHCO₃)(the molecular mass of NaHCO₃)

∴ The mass of 1.0 mol of NaHCO₃ = (1)(the molecular mass of NaHCO₃) = (the molecular mass of NaHCO₃).

∴ The mass of 1.0 mol of NaHCO₃ = (atomic weight of Na) + (atomic weight of H) + (atomic weight of C) + 3(atomic weight of O) = (23.0 g of Na) + (1.0 g of H) + (12.0 g of C) + 3(16.0 g of O) = (23.0 g of Na) + (1.0 g of H) + (12.0 g of C) + (48.0 g of O) = 84.0 g.

• To find the mass% of each element:

For Na:

The mass% of Na = (the mass of Na)/(the mass of NaHCO₃) = (23.0 g)/(84.0 g) x 100 = 27.381%.

For H:

The mass% of H = (the mass of H)/(the mass of NaHCO₃) = (1.0 g)/(84.0 g) x 100 = 1.19%.

For C:

The mass% of C = (the mass of C)/(the mass of NaHCO₃) = (12.0 g)/(84.0 g) x 100 = 14.285%.

For O:

The mass% of O = (the mass of O)/(the mass of NaHCO₃) = (48.0 g)/(84.0 g) x 100 = 57.143%.

The sum of % of elements = mass% of Na + mass% of H + mass% of C + mass% of O = 27.381% + 1.19% + 14.285% + 57.143% = 99.999% ≅ 100.0%.