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podryga [215]
2 years ago
8

What is the empirical formula of a substance that is 53.5% carbon (c), 15.5% hydrogen (h), and 31.1% nitrogen (n) by weight?

Chemistry
2 answers:
RUDIKE [14]2 years ago
8 0
Assuming 100 grams of the compound, we can convert the masses to the number of moles using the atomic masses:
     53.5 g * 1 mol C / 12.0107 g C = 4.45 mol C
     15.5 g * 1 mol H / 1.00794 g H = 15.38 mol H
     31.1 g * 1 mol N / 14.0067 g N = 2.22 mol N

We divide the number of moles of each element by the lowest number of moles, in this case, we divide by 2.22:
     C 4.45/2.22 = 2.00
     H 15.38/2.22 = 6.93 ≈ 7
     N 2.22/2.22 = 1

Therefore we can write the empirical formula using the number of moles calculated as the subscript for each element: C2H7N
Hoochie [10]2 years ago
4 0
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
since percentage compositions have been given, we can calculate the masses of elements in 100 g of the compound 
                          carbon                   hydrogen                 nitrogen 
mass                   53.5 g                     15.5 g                      31.1 g
number of moles 53.5/12                    15.5/1                     31.1 / 14 
                             = 4.46                      = 15.5                    = 2.22
divide by least number of moles 
                              4.46/2.22 = 2.0      15.5/2.22 = 6.9       2.22/2.22 = 1
numbers rounded off
C - 2
H - 7
N - 1
Then empirical formula - C₂H₇N
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irina [24]

Answer:

13.4 (w/w)% of CaCl₂ in the mixture

Explanation:

All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.

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0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻

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3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂

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Without mincing words let's dive straight into the solution to the question.

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