# Very basics general chem questions.

1. Feb 3, 2010

### Agent M27

I am in general chemistry I, and need a little clarification of the following:

When balancing an EQ say CaSO4 + AlCl3 $$\rightarrow$$ CaCl2 + Al2(SO4)2
The subscript 2 outside of the parentheses confuses me. Do I distribute that 2 to make it S2 & O8? This is a homework problem, but I do not need it solved, I just need to know what type of operation the subscript outside the parentheses will take.

Another is relating to Oxyanions. My book states the following, "Notice that when a series of oxyanions contains different numbers of oxygen atoms, they are named systematically according to the number of oxygen atoms in the ion. If there are only two ions in the series, the one with more oxygen atoms is given the ending -ate & the one with less is given the ending -ite." It then gives the example of NO3 as nitrate & NO2 as nitrite. I originally took this explanation to mean that an ion containg 3 oxygens always ends with -ate and an ion with 2 oygens ending in -ite. The question came up on a test with SO3 where and I answered with sulfate. This is infact not true, so what is the rule? It mentions a series of oxyanions, so I take it to mean NO has only two versions, where as SO has more than two versions. Is this correct? Thanks in advance everyone.

Joe

2. Feb 3, 2010

### cronxeh

It means you have 2 SO4. SO4 + SO4 has 2 S and 8 O. I'd be careful when you say O8, you have 8 oxygen atoms total on the right side, not 8 oxygen atoms bound together like in oxygen molecule (O2)

As for the second part, sulfate is SO4, sulfite is SO3, hyposulfite is SO2, and so on. No such thing as 3 oxygens 'always' being -ate. You just going to have to know this. What is certain is that lowest oxidation anion usually ends with hypo...ite and highest oxidation state ends with hyper...ate

Last edited: Feb 3, 2010
3. Feb 3, 2010

### Gokul43201

Staff Emeritus
Before you get any farther (and when you do, the answer is no, you do not distribute the subscript), it is definitely worth noting that you have an incorrect formula there (bolded). Can you figure out what the error is?