Atom, Mole, Formula, Equation

Solved Examples on the Particulate Nature of Matter

Samuel Dominic Chukwuemeka (SamDom For Peace)

Prerequisites:
(1.) Balance Chemical Reactions
(2.) Measurements and Units

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve all questions.
Show all work.

(1.) GCSE
The table shows information about two isotopes of element $X$

	Mass number	Percentage $(\%)$ abundance
Isotope $1$	$63$	$70$
Isotope $2$	$65$	$30$

$(1.1)$ Calculate the relative atomic mass $(A_r)$ of element $X$ using the equation: (***See Equation in the Answer button***)
Give your answer to $1$ decimal place.
$(1.2)$ Suggest the identity of element $X$
Use the periodic table.
$(1.3)$ The radius of an atom of element $X$ is $1.2 * 10^{-10}\;m$
The radius of the centre of the atom is $\dfrac{1}{10000}$ the radius of the atom.
Calculate the radius of the centre of an atom of element $X$
Give your answer in standard form.

$ (1.1) \\[3ex] A_r = \dfrac{(mass\;\;number * percentage)\;\;of\;\;isotope\;1 + (mass\;\;number * percentage)\;\;of\;\;isotope\;2}{100} \\[5ex] A_r = \dfrac{(63 * 70) + (65 * 30)}{100} \\[5ex] A_r = \dfrac{4410 + 1950}{100} \\[5ex] A_r = \dfrac{6360}{100} \\[5ex] A_r = 63.6 \\[3ex] (1.2) \\[3ex] The\;\;element\;\;is\;\;Copper \\[3ex] (1.3) \\[3ex] Radius\;\;of\;\;centre\;\;of\;\;atom\;X \\[3ex] = \dfrac{1}{10000} * 1.2 * 10^{-10} \\[5ex] = \dfrac{1.2 * 10^{-10}}{10^4} \\[5ex] = 1.2 * 10^{-10 - 4} \\[3ex] = 1.2 * 10^{-14}\;m $

(2.) GCSE
$(2.1)$ A company uses chlorine to produce titanium chloride from titanium dioxide.
What is the relative formula mass $(M_r)$ of titanium dioxide, $TiO_2$?
Relative atomic masses $(A_r):\;\;\; O = 16\;\;\; Ti = 48$
Tick one box

$64$	⬜
$80$	⬜
$128$	⬜
$768$	⬜

$(2.2)$ The company calculates that $500\;g$ of titanium dioxide should produce $1.2\;kg$ of titanium chloride.
However, the company finds that $500\;g$ of titanium dioxide only produces $900\;g$ of titanium chloride.
Calculate the percentage yield.

$ (2.1) \\[3ex] M_r\;\;of\;\;TiO_2 \\[3ex] = 48 + (16 * 2) \\[3ex] = 48 + 32 \\[3ex] = 80 \\[5ex] (2.2) \\[3ex] Actual\;\;Yield = 900\;g \\[3ex] Theoretical\;\;Yield = 1.2\;kg = 1.2(1000) = 1200\;g \\[3ex] Percentage\;\;Yield = \dfrac{Actual\;\;Yield}{Theoretical\;\;Yield} * 100 \\[5ex] \%\;\;Yield = \dfrac{900}{1200} * 100 \\[5ex] \%\;\;Yield = 75\% $

(3.)

$ HCl + NaOH \rightarrow NaCl + H_2O...neutralization\:\:reaction...balanced \\[3ex] 1\:mol\:\:HCl \:\:reacts\:\:with\:\:1\:\:mol\:\:NaOH \\[3ex] \dfrac{C_A * V_A}{C_B * V_B} = \dfrac{n_A}{n_B} \\[5ex] C_A = ? \\[3ex] V_A = 35\:mL \\[3ex] C_B = 0.5\:M \\[3ex] V_B = 19.6\:mL \\[3ex] n_A = 1\:mol \\[3ex] n_B = 1\:mol \\[3ex] \dfrac{C_A * 35}{0.5 * 19.6} = \dfrac{1}{1} \\[5ex] Cross\:\:Multiply \\[3ex] 1(C_A * 35) = 1(0.5 * 19.6) \\[3ex] C_A * 35 = 0.5 * 19.6 \\[3ex] C_A = \dfrac{0.5 * 19.6}{35} \\[5ex] C_A = \dfrac{9.8}{35} \\[5ex] C_A = 0.28\:M \\[3ex] $ The molarity of the $HCl$ is $0.28\:M$

(4.) GCSE An equation for the reaction is:
$NiO + C \rightarrow Ni + CO$
Calculate the percentage atom economy for the reaction to produce nickel.
Relative atomic masses $(A_r):\;\;\; C = 12\;\;\; Ni = 59$
Relative formula mass $(M_r):\;\;\; NiO = 75$
Give your answer to $3$ significant figures.

$ Desired\;\;product = Ni \\[3ex] A_r\;\;of\;\;Ni = 59 \\[3ex] Reactants = NiO\;\;and\;\;C \\[3ex] M_r\;\;of\;\;reactants = 75 + 12 = 87 \\[3ex] Atom\;\;Economy = \dfrac{M_r\;\;of\;\;desired\;\;product}{M_r\;\;of\;\;reactants} * 100 \\[5ex] Atom\;\;Economy = \dfrac{A_r\;\;of\;\;Ni}{M_r\;\;of\;\;NiO\;\;and\;\;C} * 100 \\[5ex] Atom\;\;Economy \\[3ex] = \dfrac{59}{87} * 100 \\[5ex] = 67.81609195\% \\[3ex] \approx 67.8\% \;\;(to\;\;3\;s.f) $

(5.)

$ t = 4\:years \\[3ex] Semiannual\:\:bond \rightarrow m = 2 \\[3ex] FV = \$1000 \\[3ex] BCR = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] YTM = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] BP = ? \\[3ex] BP = \dfrac{FV * BCR}{YTM} * \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] + \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[10ex] Too\:\:long...let\:\:us\:\:solve\:\:in\:\:parts \\[3ex] \underline{First\:\:Part} \\[3ex] \dfrac{FV * BCR}{YTM} \\[5ex] = \dfrac{1000 * 0.06}{0.07} \\[5ex] = \dfrac{60}{0.07} \\[5ex] = 857.142857 \\[3ex] \underline{Second\:\:Part} \\[3ex] \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] \\[10ex] mt = 2(30) = 60 \\[3ex] 1 + \dfrac{YTM}{m} = 1 + \dfrac{0.07}{2} = 1 + 0.035 = 1.035 \\[5ex] \left(1 + \dfrac{YTM}{m}\right)^{mt} = (1.035)^{60} = 7.8780909 \\[7ex] \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = \dfrac{1}{7.8780909} = 0.126934306 \\[10ex] 1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = 1 - 0.126934306 = 0.873065694 \\[10ex] \underline{Third\:\:Part} \\[3ex] \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[7ex] = \dfrac{1000}{7.8780909} \\[5ex] = 126.934306 \\[3ex] \underline{Bond\:\:Price} \\[3ex] BP = 857.142857 * 0.873065694 + 126.934306 \\[3ex] BP = 748.342023 + 126.934306 \\[3ex] BP = 875.276329 \\[3ex] BP \approx \$875.28 $

(6.)

$ 2Al_2O_3 \rightarrow 4Al + 3O_2 \\[3ex] 2\:moles\:\:of\:\:Al_2O_3 \:\:gives\:\: 4\:moles\:\:of\:\:Al \\[3ex] 1\:mole\:\:of\:\:Al_2O_3 = (27 * 2) + (16 * 3) = 54 + 48 = 102\:grams \\[3ex] \therefore 2\:moles \:\:of\:\:Al_2O_3 = 2 * 102 = 204g \\[3ex] 1\:mole \:\:of\:\:Al = 27g \\[3ex] \therefore 4\:moles \:\:of\:\:Al = 4 * 27 = 108g \\[3ex] \implies 204\:grams\:\:of\:\:Al_2O_3 \:\:gives\:\: 108\:grams\:\:of\:\:Al \\[3ex] $ Let the number of grams of $Al$ that would be obtained from $5.8$ grams of $Al_2O_3$ be $p$

Proportional Reasoning Method
$Al_2O_3$ (grams)	$Al$ (grams)
$204$	$108$
$5.8$	$p$

$ Cross\:\:Multiply \\[3ex] p * 204 = 5.8 * 108 \\[3ex] p = \dfrac{5.8 * 108}{204} \\[5ex] p = \dfrac{626.4}{204} \\[5ex] p = 3.070588235\:grams \\[3ex] p \approx 3.1\:g \\[3ex] $ About $3.1$ grams of $Al$ would be obtained from $5.8$ grams of $Al_2O_3$