lagonlon

Soal:

Carilah Berat molekul (Mr) senyawa berikut:

H₂O CO(NH₂)₂ CH₃COOH H₂SO₄ H₃PO₄ CaHPO₄ Na₂SO₄ Fe(OH)₂ PbSO₄ Ca₃(PO₄)₂ C₁₁H₁₂O₁₁

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Penyelesaian:

Jangan lupa:

Berat Molekul (BM) = Molekul relatif (Mr)

Berat Atom (BA) = Atom relatif (Ar) = nomor massa (A)

Harga Ar suatu unsur dapat dilihat dari tabel periodik

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Berat Molekul H₂O

Mr H2O	=	2ArH + ArO
	=	(2 x 1) + 16
	=	2 + 16
	=	18

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Berat Molekul CO(NH₂)₂

Mr CO(NH2)2	=	ArC + ArO + 2ArN + 4ArH
	=	12 + 16 + (2 x 14) + (4 x 1)
	=	12 + 16 + 28 + 4
	=	60

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Berat Molekul CH₃COOH

Mr CH3COOH	=	2ArC + 4ArH + 2ArO
	=	(2 x 12) + (4 x 1) + (2 x 16)
	=	24 + 4 + 32
	=	60

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Berat Molekul H₂SO₄

Mr H2SO4	=	2ArH + Ar S + 4ArO
	=	(2 x 1) + 32 + (4 x 16)
	=	2 + 32 + 64
	=	98

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Berat Molekul H₃PO₄

Mr H3PO4	=	3ArH + ArP + 4ArO
	=	(3 x 1) + 31 + (4 x 16)
	=	3 + 31 + 64
	=	98

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Berat Molekul CaHPO₄

Mr CaHPO4	=	Ar Ca + Ar H + Ar P + 4ArO
	=	40 + 1 + 31 + (4 x 16)
	=	136

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Berat Molekul Na₂SO₄

Mr Na2SO4	=	2ArNa + Ar S + 4ArO
	=	(2 x 23) + 32 + (4 x 16)
	=	46 + 32 + 64
	=	142

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Berat Molekul Fe(OH)₂

Mr Fe(OH)2	=	ArFe + 2ArO + 2ArH
	=	(2 x 56) + (2 x 16) + (2 x 1)
	=	112 + 32 + 2
	=	146

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Berat Molekul PbSO₄

Mr PbSO4	=	ArPb + ArS + 4ArO
	=	207 + 32 + (4 x 16)
	=	120 + 62 + 128
	=	303

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Berat Molekul Ca₃(PO₄)₂

Mr Ca3(PO4)2	=	3ArCa + 2ArP + 8ArO
	=	(3 x 40) + (2 x 31) + (8 x 16)
	=	120 + 62 + 128
	=	310

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Berat Molekul C₁₁H₁₂O₁₁

Mr C11H12O11	=	11ArC + 12ArH + 11ArO
	=	(11 x 12) + (12 x 1) + (11 x 16)
	=	132 + 12 + 176
	=	320

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