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4 years ago

What is the empirical formula for a compound that contains 38.77% Cl and 61.23% O?

1 answer:

3 0

W(Cl)=0.3877
w(O)=0.6123
M(Cl)=35.5 g/mol
M(O)=16.0 g/mol

ClₐOₓ
M(ClₐOₓ)=35.5a+16.0x

w(Cl)=35.5a/(35.5a+16.0x)
w(O)=16.0x/(35.5a+16.0x)

solve a system of two equations with two unknowns
35.5a/(35.5a+16.0x)=0.3877
16.0x/(35.5a+16.0x)=0.6123

a=2
x=7

Cl₂O₇ is the empirical formula

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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4

Anna71 [15]

Answer:

31.652g of Na3PO4

Explanation:

We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:

Na3PO4 will dessicate in solution as follow:

Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.

Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e

xM Na3PO4 = (1.10 x 1)/3

xM Na3PO4 = 0.367M

Therefore, the molarity of Na3PO4 is 0.367M.

Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:

Molarity of Na3PO4 = 0.367M

Volume = 525mL = 525/1000 = 0.525L

Mole of Na3PO4 =..?

Molarity = mole /Volume

0.367 = mole /0.525

Cross multiply

Mole of Na3PO4 = 0.367 x 0.525

Mole of Na3PO4 = 0.193 mole.

Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:

Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol

Mole of Na3PO4 = 0.193 mole

Mass of Na3PO4 =.?

Mass = mole x molar mass

Mass of Na3PO4 = 0.193 x 164

Mass of Na3PO4 = 31.652g

Therefore, 31.652g of Na3PO4 is needed to prepare the solution.

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