Percentage Composition, Empirical and Molecular Formulas

22.99 g (1 mol) of Na
12.01          (1 mol) of H
48.0 (1 mol) of C
48.0 (3 mole $\times$ 16.00 gram per mole) of O
The mass of one mole of $NaHCO _3$ is $ 22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g$
And the mass percentages of the elements are
mass % $Na = \dfrac{22.99 g}{84.01 g} \times 100 = 27.36$%
mass % $H = \dfrac{1.01 g}{84.01 g} \times 100 = 1.20$%
mass % $C = \dfrac{12.01 g}{84.01 g} \times 100 = 14.30$%
mass % $O = \dfrac{48.00 g}{84.01 g} \times 100 = 57.14$%

Mass of mixture=$Y-X$

mass of beaker $= 23g$
mass of mix + beaker$ = 50g$
mass of mixture $= 50 - 23 = 27g$
$%$ composition of mixture $= 35%$
Mass of the x component $= 35 \times  \dfrac{27}{100} = 9.45g$

Mass of beaker = 20 g
Mass of mixture + beaker = 56 g
Mass of mixture = 56 - 20 = 36 g
Mass of washed and dried sand = 20 g
100 g of mixture contains $= \dfrac{20}{36} \times 100 = 55$% of sand

Molar mass of compound:
Mass due to carbon:   12.01 g/mol
Total molar mass $= 12.01 + 2(16.00) = 44.01g/mol$             ($CO _2$ has two $O _2$ atoms)
Percent composition of carbon: $12.01/44.01 \times 100 = 27.28$%

$%$ $ composition =\cfrac{Mass \quad of\quad an \quad element.}{Total \quad mass \quad of\quad compound.}$

Mass of mixture=$50 g-20 g=30 g$

Total mass$=2.5+3.0g=5.5g$

Mass of mixture=$25g-10g=15g$

$PV=nRT$

Consider the mass of $C=x$

Arsenic (As) in its +3 oxidation state forms AsCl3, but in the vapor phase, it exists as a dimer, As2Cl6.

If two compounds have the same empirical formula but different molecular formula, they must have different molecular weights.