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Kazeer [188]
2 years ago
11

A compound is 52.0% zinc 9.6% carbon and 38.4% oxygen. Calculate the empirical formula of the compound.

Chemistry
1 answer:
skad [1K]2 years ago
6 0

Answer:

So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.

Explanation:

What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.

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A sample of Mo(NO3)6 has 2.22 x 10^22 nitrogen atom, how many oxygen atoms does the sample have?
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There are 3.98 × 10^23 atoms of oxygen in the sample.

Given that;

1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen

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x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms

x = 0.0368 moles

The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18

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Learn more: brainly.com/question/9743981

4 0
2 years ago
1. What amount of ammonia (in moles) is produced by the reaction of 4.00 mol H2 with 3.00 mol Nz?
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<h2>Answer:  6 moles</h2>

<h3>Explanation:</h3>

3 H₂    +    N₂   →   2 NH₃

   ↓            ↓

4 mol       3 mol

Since the moles of N₂ is the smaller of the two reactants, then N₂ is the limiting factor (the reactant that will decide how much ammonia is produced since it has the smaller amount of moles). ∴ we have to use it in calculating the number of moles of ammonia

The mole ratio of N₂ to NH₃ based on the balanced equation is 1 to 2.

∴ the moles of NH₃ = moles of N₂ × 2

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Answer:

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Explanation:

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