6,237.10²² atoms are in 1.50 g of butane
Further explanation
Stokiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.
In stoichiometry therein includes
- 1. the relative atomic mass (Ar) is a measurement of the mass of an atom relative to the mass of 1 carbon-12 atom
- 2. The relative molecular mass (M) is the sum of the mass of all the atomic masses that make up the molecule or is the sum of the relative atomic masses that make up the molecule
M X = (average mass of 1 molecule AxBy) / 1/12 mass 1 atom C-12
M AxBy = (x.Ar A + y. Ar B)
- 3. Mole is a unit of the number of particles where 1 mole is the number of particles contained in a substance (atom, molecule)which is equal to atoms in 12 gr C-12
1 mol = 6.0210²³ particles
While the number of moles can also be obtained by dividing the mass (in grams) by the relative mass of the element (Ar) or the relative molecular mass (M)
mole = gram / (Ar / M)
For example
1 mole of H element contains 6.02 × 10²³ H atoms
1 mole of water contains 6.02 × 10²³ water molecules.
1 mole of KCl ion contains 6.02 × 10²³ K⁺ ions and 6.02 × 10²³ Cl⁻ ions.
We know Butane molar mass = 58 grams / mol and mass known to be 1.5 grams so the number of moles:
mole = gram / M
mole = 1.5 / 58
mole = 0.0259
So the number of moles of C atoms in Butane is:
= 4 (valence C) x number of moles of Butane
= 4. 0.0259
= 0.1036 moles
So the number of atoms of element C is
= 0.1036 x 6.02 × 10²³
= 6,237.10 ^ 22 atoms
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Keywords: Avogadro, mole, butane
