Explore BrainMass

Explore BrainMass

Atomic Theory/Nomenclature

1) A reaction of one liter of chlorine gas with 3 liters of fluorine gas yields 2 liters of a gaseous product. All gas volumes are at the same temperature and pressure. What is the formula of the gaseous product?

2) Hydrazine, ammonia, and hydrogen azide all contain only nitrogen and hydrogen. The mass of hydrogen that combines with 1.00g of nitrogen for each compound is 1.44*10^-1g, 2.16*10^-1g and 2.40*10^-2g, respectively. Show how these data illustrate the law of multiple proportions?

3) List the noble gas elements. Which of the noble gases has only radioactive isotopes? Which lanthanide element and which transition element have only radioactive isotopes?

4) Name these compounds.
a) Hgsub2O
b) Rbsub2O
c) CaS
d) AlIsub3
e) CsF
f) Lisub3N
g) Agsub2S
h) MnOsub2
i) TiOsub2
j) Srsub3Psub2
k) BaSOsub4
l) NaNOsub2
m) KMnOsub4
n) Ksub2Crsub2Osub7
o) HCsub2Hsub2Osub2
p) NHsub4NOsub2
q) Cosub2Ssub3
r) ICl

5) Write the formula for the compounds. Side note: What is the best way to learn how to name these things?
a) sulfur difluoride
b) sulfur hexafluoride
c) sodium dihydrogen phosphate
d) lithium nitride
e) chromium III carbonate
f) tin II fluoride

Solution This solution is FREE courtesy of BrainMass!

1) You need to remember that at the same temperature and pressure, the volume of a gas is proportional to the number of molecules (or atoms for noble gases) which are in the gas. Chlorine and fluorine are both diatomic, so the ratio of their volumes tells you the ratio of the atoms in the product, and hence the formula.

2) Nitrogen (N2) has a molar mass of 28 g/mol, so 1.00 g is 3.57 x 10-2 mol.
From the molar mass of hydrogen you can similarly calculate the number of moles in 1.44*10^-1g, 2.16*10^-1g and 2.40*10^-2g. Comparing the number of moles of nitrogen to the number of moles of hydrogen gives a simple ratio in each case: the empirical formula of each compound. The law of multiple proportions is not one I'm familiar with: My guess is that it means that when substances react, they can often react in different ratios and hence produce different compounds. For example, carbon and oxygen can react to form carbon dioxide or carbon monoxide depending on the conditions.

3) The noble gas elements are the ones in the far right column of the periodic table. The transition elements are the middle eight columns of the ten-column-wide central block of the table. The lanthanides are the 14 elements starting at Lanthanum. Most periodic tables will indicate which elements exist only in radioactive states, often by showing the atomic mass in brackets e.g. (238) - indicating that there is no well-defined atomic mass for this element, but the most stable isotope has a mass number of e.g. 238.

4) For (a) and (b) there is a +1 charge on the metal atom in both cases (The oxide has a -2 charge). Rubidium, like other groups, 1 and group 2 elements, has only one oxidation state so (b) is rubidium oxide. (a) is mercury (I) oxide - the (I) showing that the mercury has, in this case, a +1 charge.

(a) to (j) and (q) and (r) are similar. Only Hg, Ag, Mn, Ti, Co and I need their oxidation numbers quoted, the other positively-charged elements are groups 1 or 2.

There are some odd ions in examples (k) to (p); some of which have both systematic and common names.
SO4-2 is sulphate (VI) or just sulphate, NO2- is nitrate (III) or nitrite, MnO4- is manganate (VII) or permanganate, Cr2O7-2 is dichromate (VI), NH4+ is ammonium

(o) is baffling: HCsub2Hsub2Osub2 can't be right. C2H3O2- is the ethanoate or acetate ion. C2O2-2 is the ethanedioate or oxalate ion. Is it one of those?

5) Phosphate is PO4-3. Dihydrogen phosphate is H2PO4-
Carbonate is CO3-2
Difluoride means there are two fluorine's in the compound, and hexafluoride means that there are six.

The best way to learn these is to practice. There are rules, but many chemists use older or simplified names. It's a lot like learning a language, there are regular bits of grammar that help, but mostly it's about learning the vocabulary and by practicing using the language.