<span>From the Born–Landé equation the lattice energy U ∝ (Z+ × Z-) / (r+ + r-) where Z+ and Z- are the charges on the cation and anion, respectively; r+ and r- are radii of the cation and anion, respectively.
The Z+×Z- term dominates. MgO has 2×2 (4) so it will have the higher U than Li2O Z+ × Z- = 2.
You only consider (r+ + r-) term when the Z+×Z- term is the same; smallest the (r+ + r-), the larger is the U.
U(MgO) = 3795 </span>kJ mol^-1
<span>U(Li2O) = 2799 kJ mol^-1
MgO has larger.
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To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.
We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'
In this, we can see that the unit 'grams' will cancel out to leave us with moles.
In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.
Answer:
atom is the smallest unit of ordinary matter that forms a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small, typically around 100 picometers across.
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
