# MolarManager.ipynb

In [1]:
from xv.chemistry.physical import MolarManager

In [2]:
ke = MolarManager()
ke

Out[2]:
3127720928496@MolarManager

Details of elements

Examples
--------
ke = MolarManager()
ke

ke.printProblemTypes()

ke.getRandomProblem()
ke.getRandomProblem(problem_type = 0)
...

ke.printProblem()
ke.printSolution()

doc_style: xv_doc

In [3]:
ke.printProblemTypes()

0. _problem_mole_in_pint
1. _problem_mole_relation
2. _problem_misc
3. _problem_mole
4. _problem_unit_to_gram
5. _problem_mole_to_gram
6. _problem_gram_to_mole

In [4]:
from IPython.display import HTML
n = len(ke._problemTemplates)
max_loop = 1
for j in range(0, max_loop):
for i in range(n):
problem_type = i
display(HTML(f"<h2>problem_type: {problem_type}/{n-1} (loop {j}/{max_loop-1})</h2>"))
ke.getRandomProblem(problem_type = problem_type, verbose = True)
display(ke.printProblem())

display(HTML(f"<h6>Solution:</h6>"))
display(ke.printSolution())
pass


## problem_type: 0/6 (loop 0/0)

Problem Template: _problem_mole_in_pint


Write relations related to mole.
###### Solution:
.Quantity'>
1.008 gram
.Quantity'>
1 mole
.Unit'>
mole
.Quantity'>
1 mole
.Quantity'>
1 mole
.Quantity'>
0.2016 mole
.Quantity'>
6.02214076e+23 / mole
.Quantity'>
6.02214076e+23 dimensionless

## problem_type: 1/6 (loop 0/0)

Problem Template: _problem_mole_relation


Write relations related to mole.
###### Solution:
1 mole = avogadro_number = 6.02214076e+23 = N

1 mole amu = 1 gram

6.022140762081123e+23 amu = 1 gram

## problem_type: 2/6 (loop 0/0)

Problem Template: _problem_misc


1. What is a mole?
a. 1 mole = 6.022e23
b. 1 mole amu = 6.022e23 amu = 1 g

2. Convert one unit the following into gram.
Hydrogen atom
Hydrogen molecule
Hydrogen ion
Oxygen atom
Oxygen molecule
Oxygen ion = (15.999 amu - 2 * 5.488e-4 amu) * ( 1 g / 6.022e23 amu )
Sodium atom
Sodium molecule
Sodium ion
H2SO4 molecule
Sulphate ion = (32.06 amu + 4 * 15.999 amu + 2 * 5.488e-4 amu) * ( 1 g / 6.022e23 amu )

3. Convert one mole of the following into gram.
Hydrogen atom
Hydrogen molecule
Hydrogen ion
Oxygen atom
Oxygen molecule
Oxygen ion
Sodium atom
Sodium molecule
Sodium ion
H2SO4 molecule
Sulphate ion

Solution for sodium ion:
1 mole = 6.022e23
1 sodium ion = (22.990 amu - 5.488e-4 amu)

1 mole sodium ion * ( 6.022e23 / 1 mole ) * ( (22.990 amu - 5.488e-4 amu) / sodium ion )

4. Convert 100 gram of the following into mole.
Hydrogen atom
Hydrogen molecule
Hydrogen ion
Oxygen atom
Oxygen molecule
Oxygen ion
Sodium atom
Sodium molecule
Sodium ion
H2SO4 molecule
Sulphate ion

Solution for Oxygen molecule
1 mole = 6.022e23
1 Oxygen molecole = 2 * 15.999 amu
100 g ( 6.022e23 amu / 1 g )
* (1 Oxygen molecule / 2 * 15.999 amu)
* ( 1 mole / 6.022e23 )
= k mole Oxygen molecule

## problem_type: 3/6 (loop 0/0)

Problem Template: _problem_mole


1. What is a mole?
a. 1 mole = 6.022e23
b. 1 mole amu = 6.022e23 amu = 1 g
###### Solution:
a. 1 mole = 6.022e23
b. 1 mole amu = 6.022e23 amu = 1 g

## problem_type: 4/6 (loop 0/0)

Problem Template: _problem_unit_to_gram


2. Convert one unit the following into gram.
a. Hydrogen atom
b. Hydrogen molecule
c. Hydrogen ion
d. Oxygen atom
e. Oxygen molecule
f. Oxygen ion
g. Sodium atom
h. Sodium molecule
i. Sodium ion
j. H2SO4 molecule
k. Sulphate ion
a. Hydrogen atom (H) = 1.6738625041514446e-24 g
b. Hydrogen molecule (H2) = 3.347725008302889e-24 g
c. Hydrogen ion (H+) = 1.6729511790102956e-24 g
d. Oxygen atom (O) = 2.6567585519760876e-23 g
e. Oxygen molecule (O2) = 5.313517103952175e-23 g
f. Oxygen ion (O--) = 2.6565762869478576e-23 g
g. Sodium atom (Na) = 3.8176685486549315e-23 g
h. Sodium molecule (Na2) = 7.635337097309863e-23 g
i. Sodium ion (Na+) = 3.817577416140816e-23 g
j. H2SO4 molecule = 1.628561939554965e-22 g
k. Sulphate ion (SO4--) = 1.5951029159747593e-22 g
###### Solution:
a. Hydrogen atom (H)
mass = $\displaystyle 1.008 amu$
mass in grams = $\displaystyle 1.008 amu \left( \frac{1 g}{6.022e23 amu} \right)$ = 1.6738625041514446e-24 g

b. Hydrogen molecule (H2)
mass = $\displaystyle (2 \cdot 1.008 amu)$
mass in grams = $\displaystyle (2 \cdot 1.008 amu) \left( \frac{1 g}{6.022e23 amu} \right)$ = 3.347725008302889e-24 g

c. Hydrogen ion (H+)
mass = $\displaystyle (1.008 amu - 0.0005488 amu)$
mass in grams = $\displaystyle (1.008 amu - 0.0005488 amu) \left( \frac{1 g}{6.022e23 amu} \right)$ = 1.6729511790102956e-24 g

d. Oxygen atom (O)
mass = $15.999 amu$
mass in grams = $\displaystyle 15.999 amu \left( \frac{1 g}{6.022e23 amu} \right)$ = 2.6567585519760876e-23 g

e. Oxygen molecule (O2)
mass = $\displaystyle (2 \cdot 15.999 amu)$
mass in grams = $\displaystyle (2 \cdot 15.999 amu) \left( \frac{1 g}{6.022e23 amu} \right)$ = 5.313517103952175e-23 g

f. Oxygen ion (O--)
mass = $(15.999 amu - 2 \cdot 0.0005488 amu)$
mass in grams = $\displaystyle (15.999 amu - 2 \cdot 0.0005488 amu) amu \left( \frac{1 g}{6.022e23 amu} \right)$ = 2.6565762869478576e-23 g

g. Sodium atom (Na)
mass = $\displaystyle 22.990 amu$
mass in grams = $\displaystyle 22.990 amu \left( \frac{1 g}{6.022e23 amu} \right)$ = 3.8176685486549315e-23 g

h. Sodium molecule (Na2)
mass = $\displaystyle (2 \cdot 22.990 amu)$
mass in grams = $\displaystyle (2 \cdot 22.990 amu) \left( \frac{1 g}{6.022e23 amu} \right)$ = 7.635337097309863e-23 g

i. Sodium ion (Na+)
mass = $\displaystyle (22.990 amu - 0.0005488 amu)$
mass in grams = $\displaystyle (22.990 amu - 0.0005488 amu) \left( \frac{1 g}{6.022e23 amu} \right)$ = 3.817577416140816e-23 g

j. H2SO4 molecule
mass = $\displaystyle (2 \cdot 1.008 amu + 32.06 amu + 4 \cdot 15.999 amu)$
mass in grams = $\displaystyle (2 \cdot 1.008 amu + 32.06 amu + 4 \cdot 15.999 amu) \left( \frac{1 g}{6.022e23 amu} \right)$ = 1.628561939554965e-22 g

k. Sulphate ion (SO4--)
mass = $\displaystyle (32.06 amu + 4 \cdot 15.999 amu + 2 \cdot 0.0005488 amu)$
mass in grams = $\displaystyle (32.06 amu + 4 \cdot 15.999 amu + 2 \cdot 0.0005488 amu) \left( \frac{1 g}{6.022e23 amu} \right)$ = 1.5951029159747593e-22 g

## problem_type: 5/6 (loop 0/0)

Problem Template: _problem_mole_to_gram


3. Convert one mole of the following into gram.
Hydrogen atom
Hydrogen molecule
Hydrogen ion
Oxygen atom
Oxygen molecule
Oxygen ion
Sodium atom
Sodium molecule
Sodium ion
H2SO4 molecule
Sulphate ion
a. Hydrogen atom (H) = 1.008 g
b. Hydrogen molecule (H2) = 2.016 g
c. Hydrogen ion (H+) = 1.0074512 g
d. Oxygen atom (O) = 15.999 g
e. Oxygen molecule (O2) = 31.998 g
f. Oxygen ion (O--) = 15.997902400000001 g
g. Sodium atom (Na) = 22.990 g
h. Sodium molecule (Na2) = 45.98 g
i. Sodium ion (Na+) = 22.989451199999998 g
j. H2SO4 molecule = 98.072 g
k. Sulphate ion (SO4--) = 96.0570976 g
###### Solution:
a. Hydrogen atom (H)
$\displaystyle 1 Hydrogen = 1.008 amu$
$\displaystyle 1 mole Hydrogen \left( \frac{6.022e23}{1 mole} \right) \left( \frac{1.008 amu}{Hydrogen} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 1.008 g$

b. Hydrogen molecule (H2)
$\displaystyle 1 H2 = (2 \cdot 1.008 amu)$
$\displaystyle 1 mole H2 \left( \frac{6.022e23}{1 mole} \right) \left( \frac{2 \cdot 1.008 amu}{H2} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 2.016 g$

c. Hydrogen ion (H+)
$\displaystyle 1 H+ = (1.008 amu - 0.0005488 amu)$
$\displaystyle 1 mole Hydrogen \space ion \left( \frac{6.022e23}{1 mole} \right) \left( \frac{1.008 amu - 0.0005488 amu}{H+}\right) \left( \frac{1 g}{6.022e23 amu} \right) = 1.0074512 g$

d. Oxygen atom (O)
$\displaystyle 1 Oxygen = 15.999 amu$
$\displaystyle 1 mole Oxygen \left( \frac{6.022e23}{1 mole} \right) \left( \frac{15.999 amu}{Oxygen} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 15.999 g$

e. Oxygen molecule (O2)
$\displaystyle 1 O2 = (2 \cdot 15.999 amu)$
$\displaystyle 1 mole O2 \left( \frac{6.022e23}{1 mole} \right) \left( \frac{2 \cdot 15.999 amu}{O2} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 31.998 g$

f. Oxygen ion (O--)
$\displaystyle 1 O-- = (15.999 amu - 2 \cdot 0.0005488 amu)$
$\displaystyle 1 mole Oxygen \space ion \left( \frac{6.022e23}{1 mole} \right) \left( \frac{15.999 amu - 2 \cdot 0.0005488 amu}{O--} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 15.997902400000001 g$

g. Sodium atom (Na)
$\displaystyle 1 Sodium = 22.990 amu$
$\displaystyle 1 mole Sodium \left( \frac{6.022e23}{1 mole} \right) \left( \frac{22.990 amu}{Sodium} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 22.990 g$

h. Sodium molecule (Na2)
$\displaystyle 1 Na2 = (2 \cdot 22.990 amu)$
$\displaystyle 1 mole Na2 \left( \frac{6.022e23}{1 mole} \right) \left( \frac{2 \cdot 22.990 amu}{Na2} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 45.98 g$

i. Sodium ion (Na+)
$\displaystyle 1 Na+ = (22.990 amu - 0.0005488 amu)$
$\displaystyle 1 mole Sodium \space ion \left( \frac{6.022e23}{1 mole} \right) \left( \frac{22.990 amu - 5.488e-4 amu}{Na+} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 22.989451199999998 g$

j. H2SO4 molecule
$\displaystyle 1 H2SO4 = (2 \cdot 1.008 amu + 32.06 amu + 4 \cdot 15.999 amu)$
$\displaystyle 1 mole H2SO4 \left( \frac{6.022e23}{1 mole} \right) \left( \frac{2 \cdot 1.008 amu + 32.06 amu + 4 \cdot 15.999 amu}{H2SO4} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 98.072 g$

k. Sulphate ion (SO4--)
$\displaystyle 1 SO4-- = (32.06 amu + 4 \cdot 15.999 amu + 2 \cdot 0.0005488 amu)$
$\displaystyle 1 mole SO4-- \left( \frac{6.022e23}{1 mole} \right) \left( \frac{32.06 amu + 4 \cdot 15.999 amu + 2 \cdot 0.0005488 amu}{SO4--} \right) \left( \frac{1 g}{6.022e23 amu} \right) = 96.0570976 g$

## problem_type: 6/6 (loop 0/0)

Problem Template: _problem_gram_to_mole


4. Convert 100 gram of the following into mole.
Hydrogen atom
Hydrogen molecule
Hydrogen ion
Oxygen atom
Oxygen molecule
Oxygen ion
Sodium atom
Sodium molecule
Sodium ion
H2SO4 molecule
Sulphate ion
a. Hydrogen atom (H) = 99.2063492063492 mole H
b. Hydrogen molecule (H2) = 49.6031746031746 mole H2
c. Hydrogen ion (H+) = 99.2603909747688 mole H+
d. Oxygen atom (O) = 6.250390649415588 mole Oxygen
e. Oxygen molecule (O2) = 3.125195324707794 mole O2
f. Oxygen ion (O--) = 6.250819482434147 mole O--
g. Sodium atom (Na) = 4.349717268377556 mole Sodium
h. Sodium molecule (Na2) = 2.174858634188778 mole Na2
i. Sodium ion (Na+) = 4.34982110403749 mole Na+
j. H2SO4 molecule = 1.0196590260216982 mole H2SO4
k. Sulphate ion (SO4--) = 1.0410474863233843 mole SO4--
###### Solution:
a. Hydrogen atom (H)
$\displaystyle 1 Hydrogen = 1.008 amu$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{Hydrogen}{1.008 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 99.2063492063492 mole H$

b. Hydrogen molecule (H2)
$\displaystyle 1 H2 = (2 \cdot 1.008 amu)$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{H2}{2 \cdot 1.008 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 49.6031746031746 mole H2$

c. Hydrogen ion (H+)
$\displaystyle 1 H+ = (1.008 amu - 0.0005488 amu)$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{H+}{1.008 amu - 0.0005488 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 99.2603909747688 mole H+$

d. Oxygen atom (O)
$\displaystyle 1 Oxygen = 15.999 amu$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{Oxygen}{15.999 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 6.250390649415588 mole Oxygen$

e. Oxygen molecule (O2)
$\displaystyle 1 O2 = (2 \cdot 15.999 amu)$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{O2}{2 \cdot 15.999 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 3.125195324707794 mole O2$

f. Oxygen ion (O--)
$\displaystyle 1 O-- = (15.999 amu - 2 \cdot 0.0005488 amu)$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{O2}{15.999 amu - 2 \cdot 0.0005488 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 6.250819482434147 mole O--$

g. Sodium atom (Na)
$\displaystyle 1 Sodium = 22.990 amu$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{Sodium}{22.990 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 4.349717268377556 mole Sodium$

h. Sodium molecule (Na2)
$\displaystyle 1 Na2 = (2 \cdot 22.990 amu)$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{Na2}{2 \cdot 22.990 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 2.174858634188778 mole Na2$

i. Sodium ion (Na+)
$\displaystyle 1 Na+ = (22.990 amu - 0.0005488 amu)$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{Na+}{22.990 amu - 0.0005488 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 4.34982110403749 mole Na+$

j. H2SO4 molecule
$\displaystyle 1 H2SO4 = (2 \cdot 1.008 amu + 32.06 amu + 4 \cdot 15.999 amu)$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{H2SO4}{2 \cdot 1.008 amu + 32.06 amu + 4 \cdot 15.999 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 1.0196590260216982 mole H2SO4$

k. Sulphate ion (SO4--)
$\displaystyle 1 SO4-- = (32.06 amu + 4 \cdot 15.999 amu + 2 \cdot 0.0005488 amu)$
$\displaystyle 100 g \left( \frac{6.022e23 amu}{1 g} \right) \left( \frac{SO4--}{32.06 amu + 4 \cdot 15.999 amu + 2 \cdot 0.0005488 amu} \right) \left( \frac{1 mole}{6.022e23} \right) = 1.0410474863233843 mole SO4--$
In [ ]: