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Elden [556K]
3 years ago
10

How many oxygen atoms are there in 582 grams of caffeine (CH10N402, 194 g/mol)? Avogadro's Number: 1 mole = 6.02 x 1023 species

A. 2 B. 6 C. 1.81 x 1024 D. 3.61 x 1024 F None of the above
Chemistry
1 answer:
Sergeu [11.5K]3 years ago
8 0

Answer:

3.61 * 10 ²⁴atoms.

Explanation:

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = 582 g

m = 194 g/mol

The number of moles can be calculated from the above formula , and substituting the respective values ,

n = w / m  = 582 g / 194 g/mol  =  3 mol

In the molecular formula of caffeine ,

<u>In 1 mole of caffeine their are - 2 moles of Oxygen. </u>

Therefore , in  3 moles of Caffeine there will be 6 moles of oxygen.

As well know ,

one mole of any substance contains 6.023*10²³ atoms,

Therefore , in  6 mol of oxygen =  6 * 6.023*10²³ atoms  = 3.61 * 10 ²⁴atoms.

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\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
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Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

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n = m/M; 0.4172 \; g of the first compound would contain

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n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
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