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2 years ago

Determine the empirical formula of a

Chemistry

1 answer:

k0ka [10]2 years ago

3 0

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold = 2.6444g

Mass of Chlorine = 0.476g

Unknown:

Empirical formula = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements Au Cl

Mass 2.6444 0.476

Molar mass 197 35.5

Number of moles 2.6444/197 0.476/35.5

0.013 0.013

Divide by the

smallest 0.013/0.013 0.013/0.013

1 1

The empirical formula of the compound is AuCl

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three moles of sodium carbonate are mixed with two moles of lead nitrate in aqueous solution, leading to formation of a solid pr

aleksandrvk [35]

There are 4 moles of spectator ions that remain in solution.

The equation of the reaction is;

Na2CO3(aq) + Pb(NO3)2(aq) -------> PbCO3(s) + 2NaNO3(aq)

We have to determine the limiting reactant. This is the reactant that yields the least amount of product. Note that the spectator ions are Na^+ and NO3^- that form NaNO3.

For Na2CO3

1 mole of Na2CO3 yields 2 moles of NaNO3

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For Pb(NO3)2

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2 moles of Pb(NO3)2 yields 2 × 2/1 = 4 moles of NaNO3

We can see that Pb(NO3)2 is the limiting reactant.

Since [NaNO3] = [Na^+] = [NO3^-], it follows that there are 4 moles of spectator ions that remain in solution.

Learn more: brainly.com/question/22885959

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3 years ago

How many mL of a 4% mass/volume Mg(NO3)2 solution would contain 1.2 grams of magnesium nitrate?

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4% mass / volume :

4 g ---------> 100 mL
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A solution has a oh- = 1 x 10-5 m what are the h30 and the ph of the solution

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Then find the pH.

pOH+pH=14

5+pH=14

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Then find the [H+] using the pH.

antilog(-9) (if you dont have an antilog button use 10^-9)

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The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the

enot [183]

Answer:

Sample B

Explanation:

In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.

Let's do Sample A:

C: 60 g; H: 12 g

1. Calculate moles:

We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):

C: 60 / 12 = 5

H: 12 / 1 = 12

2. Determine number of atoms in the formula

In this case, we just divide the lowest moles obtained in the previous part, by all the moles:

C: 5 / 5 = 1

H: 12 / 5 = 2.4 or rounded to two

3. Write the empirical formula:

Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:

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Therefore, sample A is not the same as propene.

Sample B:

C: 72 g H: 12 g

Following the same steps, let's determine the empirical formula for this sample

C: 72 / 12 = 6 ---> 6 / 6 = 1

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Sample C:

C: 84 g H: 10 g

C: 84 / 12 = 7 ----> 7 / 7 = 1

H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1

EF: CH

Sample D

C: 90 g H: 10 g

C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1

H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1

EF: CH

Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:

C3H6 -----> CH₂

Therefore, sample B is the one that match completely. Sample B would be the one.

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3 years ago