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lions [1.4K]
2 years ago
10

What is the empirical formula of a compound that is 7.74% H and 92.26% C? What is the molecular formula if the molar mass is 78.

1134 g/mol?
What is the empirical formula of a compound that is 8.68% O, 6.56% H, and 84.75% C? What is the molecular formula if the molar mass is 184.2372 g/mol?

What is the empirical formula of a compound that is 90.51% C and 9.49% H? What is the molecular formula if the molar mass is 106.167 g/mol?

What is the empirical formula and molecular formula for a substance that is 39.72% C, 1.67% H, and 58.61% Cl if the molar mass of the substance is 181.44 g/mol?

What is the empirical formula of a compound that is 47% Pt, 18.84% K, and 34.16% Cl? What is the molecular formula if the molar mass is 415.098 g/mol?

What is the empirical formula of a compound that is 6.35% H, 55.83% Cl, and 37.83% C? What is the molecular formula if the molar mass is 127.0132 g/mol?​
Chemistry
1 answer:
Minchanka [31]2 years ago
6 0

Answer:

For all these questions, we want to find the empirical and molecular formulae of various compounds given their percent composition and molar mass. The technique used to answer one of the questions can accordingly be applied to all of them.

Approaching the first question, we treat the percentages of each element as the mass of that element in a 100 g compound (as the percentages add up to 100%). So, our 100 g compound comprises 7.74 g H and 92.26 g C.

Next, we convert these mass quantities into moles. Divide the mass of each element by its molar mass:

7.74 g H/1.00794 g/mol = 7.679 mol H

92.26 g C/12.0107 g/mol = 7.681 mol C.

Then, we look for the molar quantity that's the smallest ("smaller," in this case, since there are only two), and we divide all the molar quantities by the smallest one. Here, it's a very close call, but the number of moles of H is slightly smaller than that of C. So, we divide each molar quantity by the number of moles of H:

7.679 mol H/7.679 mol H = 1

7.681 mol C/7.679 mol H ≈ 1 C/H (the value is actually slightly larger than 1, but we can treat it as 1 for our purposes).

The quotients we calculated represent the subscripts of our compound's empirical formula, which should provide the most simplified whole number ratio of the elements. So the empirical formula of our compound is C₁H₁, or just CH.

Here, it just so happens that we obtained whole number quotients. If we end up with a quotient that isn't a whole number (e.g., 1.5), we would multiply all the quotients by a common number that <em>would </em>give us the most simplified whole number ratio (so, if we had gotten 1 and 1.5, we'd multiply both by 2, and the empirical formula would have subscripts 2 and 3).

To find the molecular formula (the actual formula of our compound), we use the molar mass of the compound, 78.1134 g/mol. The molar mass of our "empirical compound," CH, is 13.0186 g/mol. Since our empirical formula represents the most simplified molar ratio of the elements, the molar masses of our "empirical compound" and the actual compound should be multiples of one another. We divide 78.1134 g/mol by 13.0176 g/mol and obtain 6. The subscripts in our molecular formula are equal to the subscripts in our empirical formula multiplied by 6.

Thus, our molecular formula is C₆H₆.

---

As mentioned before, all the questions here can be answered following the procedure used to answer the first question above. In any case, I've provided the empirical and molecular formulae for the remaining questions below for your reference.

2. Empirical formula: C₁₃H₁₂O; molecular formula: C₁₃H₁₂O

3. Empirical formula: CH; molecular formula: C₈H₈

4. Empirical formula: C₂HCl; molecular formula: C₆H₃Cl₃

5. Empirical formula: Cl₄K₂Pt; molecular formula: Cl₄K₂Pt

6. Empirical formula: C₂H₄Cl; molecular formula: C₄H₈Cl₂

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A sample of quartz is put into a calorimeter (see sketch at right) that contains of water. The quartz sample starts off at and t
pashok25 [27]

Answer:

0.71 J/g°C

Explanation:

Here is the complete question

thermometer A 51.9 g sample of quartz is put into a calorimeter (see sketch at right) that contains 300.0 g of water. The quartz sample starts off at 97.8 °C and the temperature of the water starts off at 17.0 °C. When the temperature of the water stops changing it's 19.3 °C. The pressure remains constant at 1 atm. insulated container water sample Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to 2 significant digits. a calorimeter g °C

Solution

Since the temperature of the water increases from 17.0 °C to 19.3 °C, it means that it loses heat. Also, the final temperature of the quartz equals the final temperature of the water 19.3 °C. Since the quartz temperature decreases from 97.8 °C to 19.3 °C it loses heat.

So, heat lost by quartz, Q = heat gained by water, Q'

-Q = Q'

-mc(θ₂ - θ₁) = m'c'(θ₂ - θ₃) where m = mass of quartz = 51.9 g, c = specific heat capacity of quartz, θ₁ = initial temperature of quartz = 97.8 °C, θ₂ = final temperature of quartz = 19.3 °C, m' = mass of water = 300 g, c = specific heat capacity of water = 4.2 J/g °C , θ₃ = initial temperature of water = 17.0 °C, θ₂ = final temperature of water = 19.3 °C

Making c subject of the formula, we have

c = -m'c'(θ₂ - θ₃)/m(θ₂ - θ₁)

Substituting the values of the variables into the equation, we have

c = -300 g × 4.2 J/g °C(19.3 °C - 17.0 °C)/51.9 g(19.3 °C - 97.8 °C)

c = -1260 J/°C(2.3 °C)/51.9 g(-78.5 °C)

c = -2898 J/-4074.15 g°C

c = 0.711 J/g°C

c ≅ 0.71 J/g°C to 2 significant digits

5 0
3 years ago
7. How many chlorine atoms are in a 34.2 g sample of dichlorine pentoxide?
Reptile [31]

Answer:

The answer to your question is 1.36 x 10²³ atoms

Explanation:

Data

number of atoms = ?

mass of the sample = 34.2 g

Molecule = Cl₂O₅

Process

1.- Calculate the molar mass of Cl₂O₅

Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g

2.- Calculate the atoms of Cl₂O₅

                     151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms

                       34.2 g of Cl₂O₅ ------------ x

                          x = (34.2 x 6.023 x 10²³) / 151

                          x = 1.36 x 10²³ atoms

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