what is the molar mass of carbon dioxide (co2)?

(note: c = 12.011 amu, o = 15.999 amu)

28.0 g/mol
40.0 g/mol
44.0 g/mol
58.9 g/mol

Explanation:

Atomic mass is based on a relative scale and the mass of 12C

(carbon twelve) is defined as 12 amu; so, this is an exact number.

Why do we specify 12C? We do not simply state that the mass of a C atom is

12 AMU because elements exist as a variety of isotopes.

Carbon exists as two major isotopes, 12C, and 13C ( 14C exists and has

a half life of 5730 y, 10C and 11C also exist and their half lives are

19.45 min and 20.3 days respectively). Each carbon atom has the

same number of protons and electrons, 6. 12C has 6 neutrons, 13C has

7 neutrons, and 14C has 8 neutrons and so on. So, we must specify

which C atom defines the scale.

All the masses of the elements are determined relative to 12C.

Average Atomic Mass

Since many elements have a number of isotopes, chemists use average

atomic mass. On the periodic table the mass of carbon is reported as

12.011 amu. No single carbon atom has a mass of 12.011, but in a handful

of C atoms the average mass of a carbon atom is 12.011.

Why 12.011?

If a sample of carbon was placed in a mass spectrometer the

spectrometer would detect two different C atoms, 12C and 13C.

The natural abundance of 14C, 10C and 11C in geologic (i.e. old) samples is so low that we cannot

detect the effect these isotopes have on the average mass.

From the information collected from the mass spectrometer the average

mass of a carbon atom is calculated.

The mass of 12C is, of course, 12 amu.

13C is 1.0836129 times heavier than 12C; so, the mass of 13C is 13.003355

24

amu.

98.89% of the sample is 12C, and

1.11% of the sample is 13C

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g