A g sample of H contains x H atoms.
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1 7 Chemical Composition 7. Avogadro s umber n 8, the talian scientist, Amadeo Avogadro proposed that: Equal volumes of gas at equal temperatures and pressures have the same number of particles. This law, known as Avogadro s Law, is one of the central ideas of chemistry. ne consequence of Avogadro s Law is that if a sample of an element has a mass in grams that equals the average atomic mass of the element, it will have 6.05 x 0 atoms of the element. The number, 6.05 x 0 is known as Avogadro s umber. Avogadro s umber refers to a package of particles known as a e (S unit: ) A 6.05 x 0 particles This is a kind of chemical packaging analogous to dozen refers to units. Just as you can refer to dozen eggs or dozen bags of popcorn, one can equally refer to of eggs or one e of bags of popcorn. n the latter cases the e of eggs or of bags of popcorn refer to 6.05 x 0 individual eggs or 6.05 x 0 individual bags of popcorn. The importance of Avogadro s umber is that it relates a unit mass of chemical to the number of particles of that particular chemical. n terms of atoms, one e of atoms is obtained when we weigh out exactly the average mass of the atom given in the Periodic Table, in units of grams. n other words, the number of particles in a sample can be indirectly counted by weighing out the sample. Thus, for example in the case of hydrogen, H.008 A.008 g sample of H contains 6.05 x 0 H atoms. Example: (the masses of the elements below are obtained from the Periodic Table) A sample of.0 g C contains 6.0 x 0 C atoms C atoms. A sample of 6.98 g Al contains 6.0 x 0 Al atom Al atoms. A sample of g Au contains 6.0 x 0 Au atoms Au atoms. A sample of 6.00 g contains 6.0 x 0 atoms atoms. Using Avogadro s umber, we can calculate the number of particles, and es, contained in any mass of chemical using the technique of dimensional analysis. Example:. Calculate the number of es of atoms and the number of atoms in a 3.75 g sample of carbon. C n 3.75 g C 0.33 C C.0 gc atoms C 0.33 C atoms C C C f course, if the question simply asked for the number of Carbon atoms, we could simply combine the steps: C atoms C 3.75 g C atoms C C.0 gc C page 55
2 . Calculate the number of es of atoms and the number of atoms in a 3.75 g sample of aluminum. Al n 3.75 g Al Al Al g Al atoms Al Al atoms Al Al Al or Al atoms Al 3.75 g Al atoms Al Al g Al Al 3. Calculate the number of es of atoms and the number of atoms in a 3.75 g sample of gold. Au n 3.75 gau Au Au gau atoms Au Au.70 0 atoms Au Au Au or Au atoms Au 3.75 g Au.70 0 atoms Au Au g Au Au The main point with all this is that the same mass of different materials will have different numbers of es (and particles). By knowing the mass of the particular substance, we can calculate the number of particles present. mass es particles Through Avogadro s umber and knowing the unit mass of a substance, can count the number of particles by weighing. Care must be taken however, to know exactly the type of substance we are dealing with. For example, a sample of 6.00 g contains 6.0 x 0 oxygen-atoms but only 3.0 x 0 oxygenecules. This is because the oxygen-ecule is diatomic, that is, it consists of oxygen-atoms g F H G g F g H G g atoms KJ F H G atoms K J atoms ecule KJ F H G atoms K J F H G atoms K J atoms ecules The second calculation can be done by using the packaging of es of atoms in es of ecules. F 600. g H G g KJ F H G K J F HG ecules KJ ecules page 56
3 7. Molecular Mass We have seen that we can count the number of atoms and ecules present in an element by weighing out a sample of the element and using Avogadro s umber to convert the mass to the number of atoms and ecules. The question is can we do the same when we have a ecule rather than an element? The answer is yes and this is demonstrated in the following examples. Examples:. What is the unit mass of carbon monoxide, C? Carbon monoxide consists of one atom of carbon and one atom of oxygen. Thus its mass is the sum of the masses of its component atoms. Unit mass of C.0 amu C amu 8.0 amu C Molecular Mass sum of the masses of the individual atoms This unit mass of C is known as the ecular mass (or ar mass) of C.. What is the ecular mass of methane, CH? From the Periodic Table, the atomic mass of C is.0 and of H is.008. The Molecular Mass of CH is: atom C F. 0 amu C 008. amu H atoms H HG atom C K J F + H G atom C K J. 3. What is the ecular mass of aluminum oxide, Al 3? From the Periodic Table, the atomic mass of Al is 6.98 and of H is The Molecular Mass of Al 3 is: atom Al F amu Al amu 3 atoms HG atom Al K J F + H G atom K J amu CH 096 amu Al3. What is the ecular mass of iron () sulphate, FeS? From the Periodic Table, the atomic mass of Fe is 55.85, S is 3.07 and is atom Fe F amu Al amu S amu atom S atom HG atom Al K J F + H G atom S K J F + H G atom K J. 5. What is the ecular mass of ammonium phosphate, (H ) 3 P? From the Periodic Table, the atomic mass of: H P amu FeS page 57
4 There are two main ways to go about calculating the ar mass of ionic compounds such as ammonium phosphate. i) Ammonium phosphate consists of 3 atoms, atoms H, atom P and atoms..007 amu.008 amu H amu P 6.00 amu 3 atoms + atoms H + atom P + atoms 9.09 amu ( H ) P 3 atom atom H atom P atom ii) An alternative is to calculate the mass of the individual ions and put them together in the ecule amu.008 amu H + Mass of H atom atoms H amu H atom atom H amu P 6.00 amu 3 Mass of P atom P atoms 9.97 amu P + atom P atom amu H amu P Molecular Mass ( H ) P 3 ions H 9.09 ( ) 3 ion P amu H P ion H ion P nce we have calculated the ecular mass (sometimes referred to as the ar mass) of a compound we may calculate the number of ecules in a given sample of the compound. Examples:. Calculate the number of ecules of FeS in a.37 g sample of FeS. The ar mass of FeS is 5.9. FeS ecules.37 g FeS ecules FeS FeS 5.9 g FeS. Calculate the number of ecules of (H ) 3 P in a.7 g sample of (H ) 3 P. ( H ) P ecules g H P ecules H P.7 ( H) 3P ( ) g ( H ) P 3 ( ) 3 We can also calculate the number of atoms and ecules of the elements that made up the compound. Examples:. Calculate the number of each type of atom in a 8.5 g sample of FeS. FeS n 8.5 gfes FeS FeS 5.9 gfes Fe FeS.6 0 Fe atoms Fe FeS S atoms S atoms FeS.6 0 FeS atoms FeS.58 0 atoms FeS atoms S page 58
5 . Calculate the number of each type of ecule of the element in a 8.5 g sample of FeS. FeS n 8.5 g FeS FeS FeS 5.9 gfes Fe FeS.6 0 Fe ecule Fe FeS S ecule S ecule FeS.6 0 FeS ecule FeS.9 0 ecule FeS ecule S ote that the elemental form of sulphur is S 8 but usually the monatomic form is used for simplicity. We can also calculate the mass of atoms and ecules of the elements that made up the compound. Example: Calculate the mass of each type of atom in a 8.5 g sample of FeS. FeS n 8.5 g FeS FeS FeS 5.9 gfes Fe g Fe m FeS g Fe Fe FeS Fe m FeS S FeS S g S gs S 6.00 g m FeS.85 g FeS 7.3 Percent Composition of Compounds So far the formulas of compounds have been expressed in terms of counting the atoms present in the ecule, i.e., the compounds sodium hydrogen phosphate, a HP, consists of atoms sodium, atom each hydrogen and phosphorus and atoms of oxygen. There is an older method of expressing the composition of compounds, one that predates the discovery of atoms. This method expresses the composition of compounds through the relative mass of each of the elements in the compound. The traditional way is to express the composition as a mass percentage of each component. mass of element in the compound % mass 00% total mass of the compound page 59
6 Example: Calculate the % mass of each element in sodium hydrogen phosphate, a HP. Mass of a x Mass of H x Mass of P x Mass of x Molar mass of a HP % mass a 00% 3.39% % mass H 00% 0.700% % mass P 00%.89% % mass 00% 5.08%.96 ote that % mass total % mass a + % mass H + % mass P+ % mass 3.39%+0.700%+.89%+5.08% 00.0% Example: Calculate the % mass of each element in glucose, C 6 H 6. Mass of C 6 x Mass of H x Mass of 6 x Molar mass of C 6 H % mass C 00% 0.000% % mass H 00% 6.76% % mass 00% 53.8% 80.7 To two decimal places, the total % mass sums to 00.00% as it should. ote that the mass percent of each element in a given compound is always constant, regardless of the actual mass of the sample. Thus, for example, a g sample of glucose will still consist of 0.000%C, 6.76%H and 53.8%. This fact can be used (if the % composition is known) to calculate the mass of elements that make up a particular compound of any sample size. page 60
7 Example: A sample of an organic compound has a mass of 7.9 mg. f the compound has a composition of 37.0%C,.%H,.% and 8.%, calculate the mass of each element in the sample gc mc 7.9 m 7.5 mg C gh m 7.9 m.05 mg H H g m 7.9 m 0. mg g m 7.9 m 8.70 mg Check: mg C mg H mg mg m + + +, which is correct within the limits of significant figures. page 6
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