The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.
<h3>What is an empirical formula?</h3>
It is the minimum ratio between the elements that form a compound.
- Step 1: Divide each percentage by the molar mass of the element.
C: 18.1/12.01 = 1.51
H: 2.27/1.01 = 2.25
Cl: 79.8/35.45 = 2.25
- Step 2: Divide all the numbers by the smallest one.
C; 1.51/1.51 = 1
H: 2.25/1.51 ≈ 1.5
Cl: 2.25/1.51 ≈ 1.5
- Step 3: Multiply all the numbers by 2 so all of them are whole.
C: 1 × 2 = 2
H: 1.5 × 2 = 3
Cl: 1.5 × 2 = 3
The empirical formula is C₂H₃Cl₃.
The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.
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Answer:
c
Explanation:
Nh4OH + HCL ---> NH4Cl + H3O
so ph decreases as H3O increases
and OH also decreases
Na and Cl together form the compound
NaCl2
