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: B
Explanation:
There is an atom of one type of element and then two atoms of another type of element.
Answer:
moles H₂O = 10
Explanation:
The mass of Na₂CO₃⋅xH₂O is 3.837 g and the mass of Na₂CO₃ is 1.42g
Therefore the mass of xH₂O is 3.837 - 1.42 = 2.417 g
The molar mass of Na₂CO₃ is 106 g/mol and for H₂O is 18 g/mol
The moles of Na₂CO₃ and H₂O in the sample are:
Na₂CO₃ = 1.42 / 106 = 0.01340 moles
H₂O = 2.417 / 18 = 0.1343
Now using rule of three :
1 mole of Na₂CO₃ has x moles of H₂O
0.01340 moles of Na₂CO₃ has 0.1343 moles of H₂O
x = 1 * 0.1343 / 0.01340 = 10
Answer:
you dont do anything because then you are supporting it
if u add something
Explanation:
Answer:
A. 32.06 g/mol
Explanation:
The molar mass units are always g/mol