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What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?

zubka84 [21]

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g
H: 16.3% = 16.3 g

Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

$C: \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol$

$H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol$

We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:

$C: \dfrac{6.975}{6.975} = 1$

$H: \dfrac{16.3}{6.975} \approx 2.3$

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, thus, the minimum or empirical formula found for the compound will be:

$\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark$

I hope this helps. =)

8 0

3 years ago

I need help for some of the question please,

LenKa [72]

Answer:

Number 6

Explanation:

A reflex action is an involuntary action and nearly instantaneous movement in response to a stimulus.

5 0

3 years ago

Read 2 more answers

What's the correct answer

svp [43]

The correct answer I believe is sunlight

6 0

3 years ago

Consider the following reaction:

zzz [600]

Answer:

Question 1: Both the rate of reaction of H₂O₂ and the rate of formation of H₂O are:

$6.6\cdot 10^{-3}mol/(liter.s)$

Question 2: 0.60 moles of O₂ are formed in the first 50 s of reaction.

Explanation:

The coefficients of the balanced chemical equation tells the relation of formation of the reactants into the products, which is also the relation between their rates of reaction.

The balanced chemical equation (given) is:

$2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)$

Then, in the same time that 2 moles of H₂O₂ react, 2 moles of H₂O and 1 mol of O₂ are formed.

1. Question 1.

That means that the rate of formation of O₂ is half the rate of reaction of H₂O₂ and half the rate of formation of H₂O.

Hence, if the instantaneous rate of formation of O₂ is 3.3×10⁻³moles/(liters×seconds), then the rate of reaction of H₂O₂ and the rate of formation of H₂O is twice:

$2\times 3.3\cdot 10^{-3}mol/(liter.s)=6.6\cdot 10^{-3}mol/(liter.s)$

2. Question 2.

First you must find how much the concentration has changed.

You can read the initial concentration of H₂O₂ on the graph. It is the y-intercept. It is 1.0M.

You can read the approixmate concentration of H₂O₂ at the time 50 s. It is about 0.2M.

Then the change in concentration in the first 50 s of reaction is about 1.0M - 0.2M = 0.8M.

Now you can convert the concentration 0.8M into number of moles, using the volume (1.5 liters), assuming the liquid volume does not change, and the molarity definition (molarity is the number of moles of solute in one liter of solution)

$\Delta [H_2O_2]=\Delta moles/Volume(liters)$

$0.8M=\Delta moles/1.5liter\\\\\Delta moles=0.8M\times 1.5liter=1.2mol$

Hence, 1.2 mol of H₂O₂ have reacted in the first 50s of reaction.

From the coefficients of the chemical equation, the number of moles of O₂ formed is half the number of moles of H₂O₂ that react.

Thus, 1.2mol/2 = 0.60 moles of O₂ are formed in the first 50 s of reaction.

7 0

3 years ago

Help me with this science question on the middle one..

viktelen [127]

Networks of feeding relationships is correct

8 0

3 years ago