How many atoms of nitrogen on the reactant side

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.

Minchanka [31]

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 would 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.

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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

6 0

2 years ago

Question 8 (9 points) Match the lab equipment with its purpose

Deffense [45]

Answer:

$\sf \boxed{4} \mapsto Pipet$

$\sf\boxed{7}\mapsto Test \:tube \: rack$

$\sf\boxed{3}\mapsto Test\: table$

$\sf\boxed{5}\mapsto Scoopula$

$\sf\boxed{1}\mapsto Graduated\: cylinder$

$\sf\boxed{9}\mapsto Bunsen \:burner$

$\sf\boxed{2}\mapsto Beaker$

$\sf \boxed{8}\mapsto Spot\: plate$

$\sf\boxed{6}\mapsto Goggles$

Explanation:

Pipet is used to dispense a very small amount of liquid.

Test tube rack is used to hold multiple test tubes at the same time.

Test Table is used to view chemical reactions or hold or heat small amounts of substance.

Scoopula is used to dispense chemicals from a larger container.

Graduated cylinder is used to measure volume very precisely.

Bunsen burner is used to heat objects.

Beaker is used to transport heat or store substance.

Spot plate is used to observe the color changes of small quantities of a reacting mixture.

Goggles are used to protect the eyes from flying objects or chemical splashes.

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7 0

2 years ago

What is the importance of standardization in chemistry

Anika [276]

Answer:

Standardisation is used to determine the concentration of a volumetric solution in order to achieve accurate and reliable titration results

5 0

3 years ago

Help me pls rawr uwu SSUsSY

cricket20 [7]

Help 2 what the answer for this?

7 0

2 years ago

Read 2 more answers

Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t

Vitek1552 [10]

Answer: The mass of decane produced is $1.743\times 10^2g$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

$\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol$

The chemical equation for the hydrogenation of decene follows:

$C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)$

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = $\frac{1}{1}\times 1.225=1.225mol$ of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

$1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g$

Hence, the mass of decane produced is $1.743\times 10^2g$

5 0

2 years ago