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Anna [14]
3 years ago
6

Percent composition of a compound (molar mass= 60.06) 39.99% C, 6.727% H, 53.33% O, What is the molecular formula of the compoun

d?
Chemistry
1 answer:
abruzzese [7]3 years ago
6 0

You are calculating the empirical formula of this chemical compound, which is the question with moles, molar mass, and number of moles.

We assume the total mass of this compound is 100g. Because 39.99 +6.727+53.33 =100.047is about 100g. That's the easiest way for us to do it.

First you divide the mass of carbon by its molar mass(relative formula mass)because there is a formula about moles state: number of moles=mass/molar mass.

So, 40/12 is about 3.3. Then, the RFM of H is 1, so the number of mole is 6.7/1=6.7. The RFM of O is 16, whihc means the molar mass od oxygen is 16g, so 53.3/16, which is about 3.3 too.Next, you get the number of moles in order is; 3.3, 6.7, 3.3. Now we need to look at the ratio between these numbers. 3.3 and 3.3 has a common factor of 1, so the subscript of them are both 1. next use 6.7/3.3, and the greatest common factor is 2, so the empirical formula is CH2O.

BUT CH2O is only the empirical formula. what we need to do is calculate the RFM of CH2O first.

So, 12+2+16=30.

Then we find out how many times 30 goes into 60.06; 60.06/30 is 2.002. we simplify it as 2.

Finally, all the subscript need to times 2. therefore the molecular formula is C2H4O2.

To solve this kind of  questions, there are many steps:Know what you are calculating about, it's about the MOLECULAR formula, so you need to find out the number of moles of each elements, then the empirical formula, then the molecular formula.

) Find the RFM of the element, because that is the molar mass(mass of 1 mole) of this element.

2) number of moles= mass/molar mass. use this formula to help you get the number of moles of each element in this compound

3) look at the relationship between the number of moles of each elements. find out the ratio between them.

4)use the molar mass of the whole compound to find out the molecular formula of this compounds. calculate the RFM of the compound in empirical formula form, and then divide the molar mass by the RFM.

if i got anything wrong, please tell me :)



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Answer:

Species

Explanation:

Species is the group of organisms able to interbreed and produce fertile offspring.

Let's break down each word in the question:

"Organisms" means living thing. It can be a plant or animal like we usually think of, but it also includes the really small single-celled living things like some bacteria.

"Interbreed" means to mate with each other.

"Fertile" means that the living thing can also have babies.

"Offspring" means the children that are born.

"Fertile offspring" means that the children that are made must be able to have babies of their own. For example, if a frog and a bird could interbreed, they might produce offspring (children). But, if those frog-birds cannot also have children, then frog-bird is not a species.

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2 years ago
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amm1812

Solution:

1) Separate out the half-reactions. The only issue is that there are three of them.

<span>Fe2+ ---> Fe3+ 
S2¯ ---> SO42¯ 
NO3¯ ---> NO</span>

How did I recognize there there were three equations? The basic answer is "by experience." The detailed answer is that I know the oxidation states of all the elements on EACH side of the original equation. By knowing this, I am able to determine that there were two oxidations (the Fe going +2 to +3 and the S going -2 to +6) with one reduction (the N going +5 to +2).

Notice that I also split the FeS apart rather than write one equation (with FeS on the left side). I did this for simplicity showing the three equations. I know to split the FeS apart because it has two "things" happening to it, in this case it is two oxidations.

Normally, FeS does not ionize, but I can get away with it here because I will recombine the Fe2+ with the S2¯ in the final answer. If I do everything right, I'll get a one-to-one ratio of Fe2+ to S2¯ in the final answer.

2) Balancing all half-reactions in the normal manner.

<span>Fe2+ ---> Fe3+ + e¯ 
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯ 
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O</span>

3) Equalize the electrons on each side of the half-reactions. Please note that the first two half-reactions (both oxidations) total up to nine electrons. Consequently, a factor of three is needed for the third equation, the only one shown below:

<span>3 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]</span>

Adding up the three equations will be left as an exercise for the reader. With the FeS put back together, the sum of all the coefficients (including any that are one) in the correct answer is 15.

Problem #2: CrI3 + Cl2 ---> CrO42¯ + IO4¯ + Cl¯ [basic sol.]

Solution:

Go to this video for the solution

Problem #3: Sb2S3 + Na2CO3 + C ---> Sb + Na2S + CO

Solution:

1) Remove all the spectator ions:

<span>Sb26+ + CO32- + C ---> Sb + CO</span>

Notice that I did not write Sb3+. I did this to keep the correct ratio of Sb as reactant and product. It also turns out that it will have a benefit when I select factors to multiply through some of the half-reactions. I didn't realize that until after the solution was done.

2) Separate into half-reactions:

<span>Sb26+ ---> Sb 
CO32- ---> CO 
C ---> CO</span>

3) Balance as if in acidic solution:

<span>6e¯ + Sb26+ ---> 2Sb 
2e¯ + 4H+ + CO32- ---> CO + 2H2O 
H2O + C ---> CO + 2H+ + 2e¯Could you balance in basic? I suppose, but why?</span>

4) Use a factor of three on the second half-reaction and a factor of six on the third.

<span>6e¯ + Sb26+ ---> 2Sb 
3 [2e¯ + 4H+ + CO32- ---> CO + 2H2O] 
6 [H2O + C ---> CO + 2H+ + 2e¯]The key is to think of 12 and its factors (1, 2, 3, 4, 6). You need to make the electrons equal on both sides (and there are 12 on each side when the half-reactions are added together). You get 12 H+ on each side (3 x 4 in the second and 6 x 2 in the third). You get six waters with 3 x 2 in the second and 6 x 1 in the third.Everything that needs to cancel gets canceled!</span>

5) The answer (with spectator ions added back in):

<span>Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO</span>

6) Here's a slightly different take on the solution just presented.

<span>a) Write the net ionic equation:<span>Sb26+ + CO32- + C ---> Sb + CO</span>b) Notice that charges must be balanced and that we have zero charge on the right. So, do this:<span>Sb26+ + 3CO32- + C ---> Sb + CO</span>c) Now, balance for atoms:<span>Sb26+ + 3CO32- + 6C ---> 2Sb + 9CO</span>d) Add back the sodium ions and sulfide ions to recover the molecular equation.<span>Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO</span></span>

7) Here's a discussion of a wrong answer to the above problem.

However, after reading the above wrong answer example, look at problem #10 below for an instance of having to add in a substance not included in the original reaction.

Problem #4: CrI3 + H2O2 ---> CrO42¯ + IO4¯ [basic sol.]

Solution:

1) write the half-reactions:

<span>Cr3+ ---> CrO42¯ 
I33¯ ---> IO4¯ 
H2O2 ---> H2O</span>

I wrote the iodide as I33¯ to make it easier to recombine it with the chromium ion at the end of the problem.

2) Balance as if in acidic solution:

<span>4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯ 
12H2O + I33¯ ---> 3IO4¯ + 24H+ + 24e¯ 
2e¯ + 2H+ + H2O2 ---> 2H2O</span>

I used water as the product for the hydrogen peroxide half-reaction because that gave me a half-reaction in acid solution. It will all go back to basic at the end of the problem.

3) Recover CrI3 by combining the first two half-reactions from just above:

<span>16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯</span>

4) Equalize the electrons:

<span>2 [16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯] 
27 [2e¯ + 2H+ + H2O2 ---> 2H2O]leads to:32H2O + 2CrI3 ---> 6IO4¯ + 2CrO42¯ + 64H+ + 54e¯ 
54e¯ + 54H+ + 27H2O2 ---> 54H2O</span>

5) Add the half-reactions together. Strike out (1) electrons, (2) hydrogen ion and (3) water. The result:

<span>2CrI3 + 27H2O2 ---> 2CrO42¯ + 6IO4¯ + 10H+ + 22H2O</span>

6) Add 10 hydroxides to each side. This makes 10 more waters on the right, so combine with the water alreadyon the right-hand side to make 32:

<span>2CrI3 + 27H2O2 + 10OH¯ ---> 2CrO42¯ + 6IO4¯ + 32H2O</span>



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- Absorption spectrum is a technique used to find out what makes up a sample of a substance – in other words, a chemical analysis.
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