Humboldt State University ® Department of Chemistry

Richard A. Paselk

General Chemistry
Spring 2011

Exercise: Nomenclature
© R. Paselk 2008

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*Humboldt State University* ® *Department of Chemistry*

Richard A. Paselk
	General Chemistry	Spring 2011
Exercise: Nomenclature	© R. Paselk 2008
	Supplemental Study Modules
	Use Back Button to Return to Notes

Atoms & Isotopes

The Big Picture

You can determine Z = number of protons (p) from the Periodic table.
- For atoms the charge = zero, therefore the number of electrons = number of protons.
- For ions the charge = (#p)(1+) + (#e^-)(1-)
A = p + n

For the first row, Carbon 12 (¹²C):

The superscript is A

Symbol	Z	A	p	n	e^-
¹²C		12

The Periodic table,

IIA

IIIA

IVA

VIA

VIIA

VIIIA

12.01

IIIB

IVB

VIIB

VIIIB

IIB

gives the value of Z = 6 = p, and since it is an atom with no charge, Z also = e^-

Symbol	Z	A	p	n	e^-
¹²C	6	12	6		6

Finally, the number of neutrons, n = A-Z = 12 - 6 = 6:

Symbol	Z	A	p	n	e^-
¹²C	6	12	6	6	6

For the second row, Iron (Fe)

Symbol	Z	A	p	n	e^-
Fe		56

The value of A gives the super script

Symbol	Z	A	p	n	e^-
⁵⁶Fe		56

Now looking at the Periodic table we can find Z = p = e^-

IIA

IIIA

IVA

VIA

VIIA

VIIIA

IIIB

IVB

VIIB

VIIIB

IIB

55.85

Symbol	Z	A	p	n	e^-
⁵⁶Fe	26	56	26		26

Finally, n (# neutrons) = A - Z = 30,

Symbol	Z	A	p	n	e^-
⁵⁶Fe	26	56	26	30	26

For the third row, Ruthenium(III) (Rh³⁺)

Symbol	Z	A	p	n	e^-
Rh³⁺				58

Looking at the Periodic table

IIA

IIIA

IVA

VIA

VIIA

VIIIA

IIIB

IVB

VIIB

VIIIB

IIB

101.1

gives us Z = 45 = p

Symbol	Z	A	p	n	e^-
Rh³⁺	45		45	58

Adding p + n gives A = 103

Symbol	Z	A	p	n	e^-
Rh³⁺	45	103	45	58

Finally we note that Rh³⁺ is an ion with a charge of 3+. That is it has three more protons than it does electrons. Calculating p - charge = e^-, e^- = 45 - 3 = 42,

Symbol	Z	A	p	n	e^-
Rh³⁺	45	103	45	58	42

For the fourth row, Uranium 235(IV) (²³⁵U⁴⁺)

The superscript gives us A, while the Periodic table provides Z = 92 = p

Symbol	Z	A	p	n	e^-
²³⁵U⁴⁺	92	235	92

Subtracting p from A gives the number of neutrons, n

Symbol	Z	A	p	n	e^-
²³⁵U⁴⁺	92	235	92	143

Finally we find the number of electrons, e^- = p - charge, 92 - 4 = 88

Symbol	Z	A	p	n	e^-
²³⁵U⁴⁺	92	235	92	143	88

For the fifth row, Sulfide 37 ion (³⁷S^2-)

The superscript gives us A, while the Periodic table provides Z = 16 = p

Symbol	Z	A	p	n	e^-
³⁷S^2-	16	37	16

Subtracting p from A gives the number of neutrons, n

Symbol	Z	A	p	n	e^-
³⁷S^2-	16	37	16	21

Finally we find the number of electrons, e^- = p - charge, 16 - (-2) = 18

Symbol	Z	A	p	n	e^-
³⁷S^2-	16	37	16	21	18

For the fifth row, Arsenide 75 ion (⁷⁵As^3-)

The superscript gives us A, while the Periodic table provides Z = 16 = p

Symbol	Z	A	p	n	e^-
⁷⁵As^3-	33	75	33

Subtracting p from A gives the number of neutrons, n

Symbol	Z	A	p	n	e^-
⁷⁵As^3-	33	75	33	42

Finally we find the number of electrons, e^- = p - charge, 33 - (-3) = 30

Symbol	Z	A	p	n	e^-
⁷⁵As^3-	33	75	33	42	36

So for the whole thing:

Symbol	Z	A	p	n	e^-
¹²C	6	12	6	6	6
⁵⁶Fe	26	56	26	30	26
Rh³⁺	45	103	45	58	42
²³⁵U⁴⁺	92	235	92	143	88
³⁷S^2-	16	37	16	21	18
⁷⁵As^3-	33	75	33	42	36

Return to Atoms & Isotopes Exercises

© R A Paselk

Last modified 23 September 2008

Humboldt State University ® Department of Chemistry

Richard A. Paselk

Supplemental Study Modules

Atoms & Isotopes

For the first row, Carbon 12 (12C):

For the second row, Iron (Fe)

For the third row, Ruthenium(III) (Rh3+)

For the fourth row, Uranium 235(IV) (235U4+)

For the fifth row, Sulfide 37 ion (37S2-)

For the fifth row, Arsenide 75 ion (75As3-)

So for the whole thing:

For the first row, Carbon 12 (¹²C):

For the third row, Ruthenium(III) (Rh³⁺)

For the fourth row, Uranium 235(IV) (²³⁵U⁴⁺)

For the fifth row, Sulfide 37 ion (³⁷S^2-)

For the fifth row, Arsenide 75 ion (⁷⁵As^3-)