Explanation:
The scientists realized that the properties of elements related to each other depending on relational properties of their atoms. They realized that atoms of different elements shared some properties depending on the number of electrons they have in their valence orbital shells and atomic mass numbers. These atomic properties determined the elements' shared properties such as texture, reactivity, electric conductivity and etcetera. Arranging the atoms in a periodic table would therefore ideally make it easier to decipher the shared similarities and differences between atoms of different elements, such as physical and chemical properties.
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The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus 
= 0.0293 mol
Moles bromine 6.99 g bromine
=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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Answer:
4.43 g of Oxygen
Explanation:
As shown in Chemical Formula, one mole of Aluminium Sulfate [Al₂(SO₄)₃] contains;
2 Moles of Aluminium
3 Moles of Sulfur
12 Moles of Oxygen
Also, the Molar Mass of Aluminium Sulfate is 342.15 g/mol. It means,
342.15 g ( 1 mole) of Al₂(SO₄)₃ contains = 192 g (12 mole) of O
So,
7.9 g of Al₂(SO₄)₃ will contain = X g of O
Solving for X,
X = (7.9 g × 192 g) ÷ 342.15 g
X = 4.43 g of Oxygen
