## Sig Fig's in Problem E.21.d

Sarah_Kang_2K
Posts: 72
Joined: Fri Sep 28, 2018 12:23 am

### Sig Fig's in Problem E.21.d

On Question E.21.d of the 6th edition textbook, the first step is to calculate the molar mass of glucose (C6H12O6).
I am confused as to why the final molar mass contains 5 sig fig's and not 4.

Here is my work:

6(12.011) + 12(1.008) + 6(15.999) = 180.2 g/mol

The textbook solutions states that the answer is 180.15 g/mol

When adding numbers, aren't we supposed to use the least # of sig fig's contained in any of the 3 numbers? Using that logic, the molar mass of Hydrogen (1.008) has 4 sig fig's, so I thought that the final answer should also have 4 sig fig's.

Can someone explain to me why the final molar mass (180.15 g/mol) contains 5 sig fig's, and not 4?

shaunajava2e
Posts: 66
Joined: Fri Sep 28, 2018 12:26 am

### Re: Sig Fig's in Problem E.21.d

I'm assuming they are using the molar mass of H as its shown in the front cover of the book, in which case it would be 1.0079 g.mol-1 which has 5 sigfigs

kimberlyrose1G
Posts: 44
Joined: Fri Sep 28, 2018 12:27 am

### Re: Sig Fig's in Problem E.21.d

For most periodic tables, the mass of hydrogen is given as 1.0079, with five significant figures so the final answer would be given in a value of five significant figures, following the sig figs rules.

Chloe Likwong 2K
Posts: 70
Joined: Fri Sep 28, 2018 12:23 am

### Re: Sig Fig's in Problem E.21.d

The two precedent posts make sense, however, when one adds or subtracts, the rule for significant figures differs from the multiplication/division. When one is adding or subtracting, they should take the "smallest amount of decimal places in the data" (this can be found in one of the links in the Chem website : https://lavelle.chem.ucla.edu/wp-conten ... OUT_SF.pdf ).

For example: 77.4 + 1.35 + 2.105 = 80.855 = 80.9.

Therefore, it completely depends on how specific one wants to be in terms of rounding the masses of the atoms. According to your work, there should be 3 decimal places; however, this answer also contradicts the Solutions Manual's answer.