Chemistry HELP pleaase Im having a test soon?

A

Andy

New member
empirical formula is the mole to mole ratio of elements

1) potassium oxalate
those %s add up to be 100%
so we have to assume those %s to be grams
C= 14.45g
K=47.05g
O=38.50 g
find their moles using the mole formula
mole= give mass(g)/ gram formula mass
C= 14.45/12
C= 1.2 mole <<<<
K=47.05/ 39.1
K=1.2 mole <<<<
O=38.5/16
O= 2.4 mole <<<<

after finding the moles of each of the elements
we have to make them whole numbers
to do that we will just divide each by the smallest mole which is 1.2
C=1.2/1.2 = 1
K=1.2/1.2 = 1
O=2.4/1.2 = 2
these are the mole to mole ratios of the elements 1 : 1 : 2
the last steps is to multiply these numbers to its element
C x 1 = C
K x 1 = K
Ox 2 = O2
therefore the empirical formula is KCO2

2) nitrogen oxide = 1.52 g
oxygen = 0.96g
to get the mass of nitrogen just simply subtract the numbers
1.52g - 0.96g = 0.56g of nitrogen

do the same things as i said in question 1
find mole
divide by smallest mole if is not whole number
times it by the element

sorry have to go. couldnt help u with the rest..
 
my teacher gave me this packet, tools of chemistry some of the material he hasnt gone over but I wana know so that way when he gets on subject i can stay on track with him.

!) Find empirical formula of potassium oxalate. ( 14.45% C, 47.05%K, 38.50%O) << I dont know what these percents have to do w/anything.>>

2) another empirical formula only this problem is like 1.52g a sample of nitrogen oxide was found to contain 0.96g of oxygen....

3) what is the percent phosphorus in a laundry detergent that contains 46% by mass of water softtener Na5P3O10 as its only source of phosphorus...

4)When 0.387g of Cr is heated in an atmosphere of Cl2 gas, a combination reaction occurs and 1.178g of a solid compound is formed. Assuming that all the chromium reacts what is the empirical formula???


Thanks everyone, besides if u dont know the answer its okay, he wont be grading it, its on scantron so he scans it and hands it to me, that way i wanted to know the steps because that last packet he gave he didnt go over it.
 
Ok first question.
To find the empirical formula given the percentages.. first turn the percentages into grams by assuming that the mass is out of 100 grams.

So you would get 14.45 grams of C, 47.55g of K, and 38.40g O. Then you would multiply each of those grams by the molar mass of the element.

14.45/12.01(amu C)= 1.2 moles
47.05/39.10(amu K)= 1.2 moles
38.40g/16(amu O)=2.4 moles

From those moles you find the lowest moles which could be oxygen or carbon. Ill choose carbon. Take the moles of carbon and divide each moles by 1.2

1.2 moles C/ 1.2 moles C =1
1.2 moles K/ 1.2 Moles C =1
2.4 Moles O/ 1.2 Moles C = 2

There is the empirical formula K(1)C(1)O(2)

Umm the rest i forgot what to do ahhah
 
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