# Moles in a Cat?

1. Sep 6, 2004

I have to approximately figure out, to the nearest order of magnitude, how many moles of atoms there are in a 12 kg cat.You are also told that The masses of a hydrogen atom, an oxygen atom, and a carbon atom are 1.0 u, 16 u, and 12 u. .

all i know is that 1 u=1.66 x 10^-27 kg.
From here I am unsure where to go.

2. Sep 6, 2004

### Nylex

1 mole contains an Avogadro's amount of atoms (Avogadro's no = 6.022 x 10^23), that should help you a bit.

3. Sep 6, 2004

Yes I knew that before working on this problem and yet I'm still lost.

4. Sep 6, 2004

### faust9

In what proportion are the main ingrediants found in the cat (ie cat's are 80% water). If you know the mass the proportion of the three elements then you should eb able to find number of moles of atoms.

5. Sep 6, 2004

I have no other information besides what I wrote before.

6. Sep 6, 2004

### faust9

Then make some assumptions or do some research. How much water is in a cat? Once you know that you can 'assume' the rest or a portion of the rest is carbon. Once you know how much water is in a cat and the mass of the cat you can easily calculate the moles og H and O. Once you make your assumption about C then you'll be able to calculate the number of C atoms.

7. Sep 8, 2004

### 0aNoMaLi7

http://www.sniksnak.com/cathealth/howto9.html

The above link says: "...Water comprises from about 95 percent of the new-born kitten to about 75 percent of the adult cat..."

So you have H and O....maybe an estimate is in order?

Im having difficulty with the same problem ;-). I PMed you GingerBread27 with a question FYI :-). Best of luck.

8. Sep 8, 2004

### 0aNoMaLi7

k, i just did the problem myself....follow me...

H= 1.0u
O= 16.0u
C= 12.0u

we don't know the age of the kitten so lets say roughtly 80% of the given cat is water

in my version of the problem, my cat is 5 kg....therefore

80% x 5 kg = 4 kg = amount of H20

convert to grams.....(habit from chemistry)

4kg = 4000g

4000gH20 x (1 mol H20/ 18g H20) = 222 mol H20

we have 1 kg of the 'cat' unallocated therefore let the remaining carbon equal the 1 kg

1000gC x (1 mol C / 12g C) = 83.3 mol C

total 'molage' => 222+83 = 305 mol

and i was told by my school online software that this was correct. hope this helps...if i made a mistake i apologize - its 1:40 am! Cheerz.