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3 years ago

What is the molecular formula for a compound that is 44.87% potassium, 36.7%

Chemistry

1 answer:

Phoenix [80]3 years ago

5 0

Answer:

Molecular formula: S4K8O16 empirical formula: SK2O4

Explanation:

First we find the moles of each by first finding grams (using the percent) and then using stoichiometry to convert into moles:

Sulfur: 696 *.18 = 125.28grams S* $\frac{1 mole S}{32.065 g S} = 3.907 moles S$

Potassium: 696 *.4487 = 312.2952 * $\frac{1 mole K}{39.08 g K}$ = 7.99117 mole K

Oxygen: 696 * .367 = 255.432 * $\frac{1 mol O}{16g O}$ = 15.9654 mole O

Then we divide each value by the atom with the smallest number of moles to find the mole ratio:

3.907/3.907= 1

7.99117 mole K/ 3.907= 2.043

15.9654 mole O/ 3.907= 4.08

The empirical formula is SK2O4

To find the molecular formula, we divide the mass given (696) by the mass of the empirical formula (174.22) to get 4. We then divide each atom by 4.

Molecular formula: S4K8O16

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3 years ago

Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu

Deffense [45]

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

$r = \dfrac{a}{2\sqrt{2}}$

where;

a = edge length.

Making "a" the subject of the formula:

$a = 2 \sqrt{2} \times r$

$a = 2 \times 1.414 \times 135 \ pm$

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

$density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}$

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

$mass\ of\ atom =\dfrac{ atomic \ mass}{Avogadro \ number}$

$mass\ of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}$

Recall also that number of atoms in a unit cell for a face-centered cubic = 4

So;

$mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4$

mass of unit cell = 6.83380375 × 10⁻²² g

$Density \ of \ unit \ cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}$

Density of unit cell ( rhodium) = 12.279 g/cm³

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3 years ago

Electron transfer is the term for this process. The resulting anion and cation are attracted by Coulombic forces and an ionic co

Nitella [24]

<h2>The compound formed is NaCl and $CaCl_{2}$ </h2>

Explanation:

PART 1:

Atomic number of sodium is 11.
Sodium(Na, Z = 11, Group 1A) will lose 1 electron to become $Na^{1+}$ which is isoelectronic to Neon (Ne ,Z = 10).
The symbol for the sodium ion is $Na^{+}$ .
Chlorine atom ( $Cl$ , Z = 17, Group 7A or 17) gains 1 elecron to be isoelectronic to neon.
The symbol for the compound formed is NaCl.
PART 2:
Atomic number of calcium is 20
Calcium(Ca, Z = 20 ,Group 2A) will lose two electrons to become $Ca^{2+}$ which is isoelectronic to argon.
The symbol for the ion is $Ca^{2+}.$
Two chlorine atoms (Cl, Z = 17, Group 7A or 17) each gains one electron to be isoelectronic to argon(Z = 18)
The symbol for the compound formed is $CaCl_{2}$ .